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Conic Sections: Chapter Test

20 multiple-choice questions and 10 free-response questions, drawn from across the chapter and mixed together.

Free response

10 questions in parts, 111 points in total. Work them out on paper. There are no hints here: reveal each question's answer, worked solution, and rubric when you are ready to mark that one.

Free response · work it on paper

Question 1 of 10

1. One equation, read two ways from two given lines . 12 points. Question 1 of 10.

A parabola has focus $(-3, 4)$ and directrix $x = 5$ .
1. Part A.
  A point $P=(x,y)$ belongs to this parabola exactly when $PF = \operatorname{dist}(P, d)$ . Turn that single sentence into an equation, and simplify what you get to the form $(y-k)^2 = 4p(x-h)$ .
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
2. Part B.
  From your equation in part A, read off the vertex, $p$ , and the direction the parabola opens. Confirm these give back the stated focus and directrix.
  
  Carry your own answer forward Read these values off whatever equation you reached in part A; credit is for reading $h$ , $k$ , and $p$ off your own coefficients correctly, not for matching one particular equation.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  Without repeating the full derivation, explain in general (using letters, not the specific numbers above) why $4p$ , and not $p$ alone, is always the coefficient produced when this squaring process is carried out on a focus $(h+p,k)$ and directrix $x=h-p$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer
Part A

$(y-4)^2 = -16(x-1)$ .
- Any equivalent unsimplified line reached before it is written in the required form, such as $(y-4)^2 = -16x+16$
Part B

Vertex $(1,4)$ , $p=-4$ , opens left. Focus $(1-4,4)=(-3,4)$ and directrix $x=1-(-4)=5$ match the given data.

Part C

Writing $u=x-h$ , the squared equation reduces to $(u-p)^2+(y-k)^2=(u+p)^2$ , and expanding both squares leaves $(y-k)^2=(u+p)^2-(u-p)^2=4up=4p(x-h)$ : the factor of $4$ comes from the difference of these two squares, not from any choice specific to one parabola.
Worked solution

Part A

Write both distances as expressions in $x$ and $y$ and set them equal.

$\sqrt{(x+3)^2+(y-4)^2} = |x-5|.$

Both sides are nonnegative, so squaring is reversible.

$(x+3)^2+(y-4)^2=(x-5)^2.$

Expand the two squares in $x$ : $(x+3)^2=x^2+6x+9$ and $(x-5)^2=x^2-10x+25$ .

$x^2+6x+9+(y-4)^2=x^2-10x+25.$

The $x^2$ terms cancel, leaving

$(y-4)^2=-10x+25-6x-9=-16x+16=-16(x-1).$

Part B

Match $(y-4)^2=-16(x-1)$ against $(y-k)^2=4p(x-h)$ : vertex $(h,k)=(1,4)$ , and $4p=-16 \;\Longrightarrow\; p=-4$ . Since $p<0$ , the parabola opens toward decreasing $x$ , i.e. left.

$\text{focus}=(h+p,k)=(1-4,4)=(-3,4), \qquad \text{directrix}=x=h-p=1-(-4)=5.$

Both match the values given at the start.

Part C

Repeat the setup with letters: focus $(h+p,k)$ , directrix $x=h-p$ , point $(x,y)$ . Let $u=x-h$ .

$\sqrt{(u-p)^2+(y-k)^2}=|u+p| \;\Longrightarrow\; (u-p)^2+(y-k)^2=(u+p)^2.$

Expand both squares in $u$ : $(u-p)^2=u^2-2up+p^2$ and $(u+p)^2=u^2+2up+p^2$ . The $u^2$ and $p^2$ terms cancel from both sides.

$(y-k)^2=(u+p)^2-(u-p)^2=4up=4p(x-h).$

The factor of $4$ arises specifically from the difference of two squares, $(u+p)^2-(u-p)^2$ , which always equals $4up$ no matter what $p$ or $h$ are. So $4p$ , never $p$ alone, is guaranteed to be the coefficient for every parabola built this way.

In one line

$(y-4)^2=-16(x-1)$ , vertex $(1,4)$ , $p=-4$ , opening left, with focus $(-3,4)$ and directrix $x=5$ exactly as given. In general, expanding $(u+p)^2-(u-p)^2=4up$ shows the coefficient is always $4p$ , never $p$ alone, for any parabola built this way.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the distance to the focus and the distance to the directrix as expressions in $x$ and $y$ , and sets them equal before squaring. . Worth 2 points.

Squares both sides, expands correctly, and simplifies to isolate $(y-4)^2$ , showing the cancellation of the $x^2$ terms. . Worth 2 points. needs an explanation, not just an answer

Part B 4 points

Reads the vertex $(1,4)$ and computes $p=-4$ correctly from $4p=-16$ , not from $-16$ directly. . Worth 2 points.

States the parabola opens left, tying that to the sign of $p$ . . Worth 1 point.

Recovers the focus and directrix and confirms they match the stem. . Worth 1 point.

Part C 4 points

Redoes the derivation with general letters ( $u=x-h$ , focus $h+p$ , directrix $h-p$ ) rather than appealing only to the specific numbers above. . Worth 2 points. needs an explanation, not just an answer

Identifies that the coefficient $4p$ comes from the difference of squares $(u+p)^2-(u-p)^2=4up$ , and states that this holds for every choice of $p$ and $h$ . . Worth 2 points. needs an explanation, not just an answer
2. A range, a general form, and a shortcut worth questioning . 12 points. Question 2 of 10.

A lighthouse's warning beacon is mounted at $(2, -5)$ on a nautical chart (units in kilometers), and its beam fades out at exactly $7$ kilometers.
1. Part A.
  Model the outer edge of the beam: for a general point $(x,y)$ on that edge, its distance to the beacon equals the beam's reach. Square that relationship and multiply it out into the general form $x^2+y^2+Dx+Ey+F=0$ .
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  A fishing boat is charted at $(6, -2)$ . Plug those coordinates into the left side of your part A equation, and decide, from the sign alone, whether the boat is inside, on, or outside the beam's reach.
  
  Carry your own answer forward Use your own equation from part A, whatever its coefficients came out to be; credit is for substituting correctly and reading the sign, not for matching one particular number.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  A colleague argues that reading the sign directly from the general-form expression is a shortcut that skips a step, and that converting to center-radius form first and computing the actual distance to the center would be more reliable. Compare the two methods: is the sign-test method skipping any real work, or performing the same comparison in a different order? Justify your answer using the relationship between the general-form expression and the squared distance from the center.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 4 points
The answer

Part A

$x^2+y^2-4x+10y-20=0$ .

Part B

The boat is inside the beam's reach: plugging in gives $-24$ , a negative value.

Part C

The two methods do the same comparison. The general-form expression equals (squared distance from the center) minus $r^2$ , so evaluating its sign at a point is not a shortcut that skips anything; it performs the identical comparison without first extracting $h$ , $k$ , and $r$ explicitly.

Worked solution

Part A

The distance from the beacon $(2,-5)$ to a point on the edge equals $7$ .

$(x-2)^2+(y+5)^2=49.$

Expand and collect.

$x^2-4x+4+y^2+10y+25=49 \;\Longrightarrow\; x^2+y^2-4x+10y-20=0.$

Part B

Plug $x=6$ , $y=-2$ into the left side.

$6^2+(-2)^2-4(6)+10(-2)-20=36+4-24-20-20=-24.$

This is negative, so the boat's squared distance from the beacon is less than $r^2$ : it is inside the beam's reach.

Part C

For any circle, $x^2+y^2+Dx+Ey+F$ is exactly $(x-h)^2+(y-k)^2-r^2$ in expanded form, where the center and radius came from completing the square.

$x^2+y^2+Dx+Ey+F \;=\; (x-h)^2+(y-k)^2-r^2.$

Evaluating the general-form expression at a point $(x,y)$ therefore evaluates $(x-h)^2+(y-k)^2-r^2$ at that point, whether or not $h$ , $k$ , and $r$ were ever written down separately.

So the colleague's proposed method, converting to center-radius form and comparing $(x-h)^2+(y-k)^2$ to $r^2$ directly, computes the same quantity, term for term, as substituting into the general-form expression and comparing to $0$ . Neither method skips a step the other performs; they differ only in whether the subtraction happens before or after the center and radius are extracted by name.

In one line

The beam's edge is $x^2+y^2-4x+10y-20=0$ ; plugging in the boat's location $(6,-2)$ gives $-24$ , so the boat is inside the beam's reach. The sign-test method is not a shortcut that skips work: the general-form expression is algebraically the squared distance from the beacon minus $r^2$ , so it performs the same comparison a center-radius computation would, only without extracting $h$ , $k$ , $r$ by name first.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Writes the distance condition $(x-2)^2+(y+5)^2=49$ with correct signs on each coordinate. . Worth 2 points.

Expands both squares correctly and collects every term onto one side to reach general form. . Worth 2 points.

Part B 4 points

Substitutes $(6,-2)$ into the general-form expression from part A correctly and computes the resulting value. . Worth 2 points.

Reads the sign correctly to reach a location verdict. . Worth 2 points.

Part C 4 points

Identifies that the general-form expression equals the squared distance from the center minus $r^2$ , so its sign carries the same information as the direct distance comparison. . Worth 3 points. needs an explanation, not just an answer

Concludes explicitly that neither method skips real work; they perform the same comparison in a different order. . Worth 1 point.
3. Two ellipses, read cold, and compared without a formula . 10 points. Question 3 of 10.

Two ellipses appear in the same design brief: $\dfrac{x^2}{169}+\dfrac{y^2}{25}=1$ and $\dfrac{x^2}{40}+\dfrac{y^2}{49}=1$ .
1. Part A.
  For $\dfrac{x^2}{169}+\dfrac{y^2}{25}=1$ , find $a$ , $b$ , $c$ , state the major axis, and give the vertices and foci.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  For $\dfrac{x^2}{40}+\dfrac{y^2}{49}=1$ , find $a$ , $b$ , $c$ , state the major axis, and give the vertices and foci.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  Without computing an eccentricity, say which of the two ellipses above is more elongated (farther from circular), and justify your answer using only the RATIO between $a$ and $b$ you already found in parts A and B, not their raw difference.
  
  Carry your own answer forward Compare whichever $a$ and $b$ you found in parts A and B; the credit is for reasoning from the RATIO $b/a$ , not for matching a particular value of it.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$a=13$ , $b=5$ , $c=12$ ; major axis horizontal. Vertices $(\pm13,0)$ , foci $(\pm12,0)$ .

Part B

$a=7$ , $b=2\sqrt{10}$ , $c=3$ ; major axis vertical. Vertices $(0,\pm7)$ , foci $(0,\pm3)$ .

Part C

The first ellipse is more elongated: its ratio $b/a=5/13\approx0.385$ sits far below $1$ , while the second's $b/a=2\sqrt{10}/7\approx0.904$ sits close to $1$ , the near-circular case.

Worked solution

Part A

The larger denominator is $169$ , under $x^2$ , so the major axis is horizontal.

$a^2=169,\; b^2=25 \;\Longrightarrow\; a=13,\, b=5, \qquad c^2=169-25=144 \;\Longrightarrow\; c=12.$

Vertices $(\pm13,0)$ , foci $(\pm12,0)$ .

Part B

The larger denominator is $49$ , under $y^2$ , so the major axis is vertical.

$a^2=49,\; b^2=40 \;\Longrightarrow\; a=7,\, b=\sqrt{40}=2\sqrt{10}, \qquad c^2=49-40=9 \;\Longrightarrow\; c=3.$

Vertices $(0,\pm7)$ , foci $(0,\pm3)$ .

Part C

An ellipse's shape, unlike its size, is controlled by a RATIO, not a raw difference: stretching every length of an ellipse by the same factor changes $a-b$ but leaves the ellipse's shape (and $b/a$ ) unchanged, so $a-b$ cannot be what decides how close a curve is to a circle. The ratio $b/a$ is scale-invariant, close to $1$ exactly when the curve is close to circular ( $a\approx b$ ) and close to $0$ when it is flattened.

$\text{first: } \frac{b}{a}=\frac{5}{13}\approx0.385, \qquad \text{second: } \frac{b}{a}=\frac{2\sqrt{10}}{7}\approx0.904.$

The first ellipse's ratio sits far below $1$ , so it is the more elongated of the two, while the second's ratio sits close to $1$ , so it is close to circular.

In one line

$\dfrac{x^2}{169}+\dfrac{y^2}{25}=1$ : $a=13$ , $b=5$ , $c=12$ , major axis horizontal, vertices $(\pm13,0)$ , foci $(\pm12,0)$ . $\dfrac{x^2}{40}+\dfrac{y^2}{49}=1$ : $a=7$ , $b=2\sqrt{10}$ , $c=3$ , major axis vertical, vertices $(0,\pm7)$ , foci $(0,\pm3)$ . The first ellipse is far more elongated, since its ratio $b/a\approx0.385$ sits well below $1$ , while the second's ratio $b/a\approx0.904$ sits close to $1$ , the near-circular case.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the larger denominator as $a^2$ and the major axis as running along that variable. . Worth 1 point.

Computes $c$ by subtraction, and reports the vertices and foci on the correct axis. . Worth 2 points.

Part B 3 points

Identifies the larger denominator as $a^2$ and the major axis as running along that variable, independently of part A's orientation. . Worth 1 point.

Computes $c$ correctly and simplifies $b=\sqrt{40}$ to $2\sqrt{10}$ , and reports the vertices and foci on the correct axis. . Worth 2 points.

Part C 4 points

Identifies elongation with the RATIO $b/a$ (close to $1$ near-circular, close to $0$ flattened) rather than the raw difference $a-b$ , and explains why a ratio, not a difference, is the right measure since it does not change under uniform scaling. . Worth 2 points. needs an explanation, not just an answer

Computes the two specific ratios (about $0.385$ versus about $0.904$ ) and reaches the correct verdict that the first ellipse is more elongated. . Worth 2 points.
4. A hyperbola, read plainly, and a rule tested against its own exception . 10 points. Question 4 of 10.

A hyperbola has equation $\dfrac{x^2}{361}-\dfrac{y^2}{400}=1$ .
1. Part A.
  Find $a$ , $b$ , and $c$ , and note which of the three comes out largest.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  State the vertices, the foci, and the equations of the two asymptotes.
  
  Carry your own answer forward Use your own $a$ , $b$ , $c$ from part A.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
3. Part C.
  A classmate says that since $400>361$ , this hyperbola must open up and down, quoting the ellipse's rule that the larger denominator names $a^2$ . Explain specifically why that rule does not apply here, and state the correct test for a hyperbola's orientation.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$a=19$ , $b=20$ , $c=\sqrt{761}\approx27.6$ , the largest of the three.

Part B

Vertices $(\pm19,0)$ ; foci $(\pm\sqrt{761},0)$ ; asymptotes $y=\pm\dfrac{20}{19}x$ .

Part C

A hyperbola's orientation is decided by which term is POSITIVE, not which denominator is larger; here $x^2$ is positive even though its denominator ( $361$ ) is smaller than $y^2$ 's ( $400$ ), so the branches open left and right despite $400>361$ .

Worked solution

Part A

The $x^2$ term is positive, so $a^2=361$ and $b^2=400$ .

$a=19, \quad b=20, \qquad c^2=a^2+b^2=361+400=761 \;\Longrightarrow\; c=\sqrt{761}\approx27.6.$

Since $c^2=a^2+b^2$ adds two positive numbers, $c$ is always the largest of the three for a hyperbola.

Part B

Vertices and foci lie on the transverse (horizontal) axis, $a$ and $c$ from the center: $(\pm19,0)$ and $(\pm\sqrt{761},0)$ .

The asymptote slope is $\pm\frac{b}{a}$ when $x^2$ is positive.

$y=\pm\frac{20}{19}x.$

Part C

The ellipse's rule (larger denominator names $a^2$ , and the major axis follows it) comes from the requirement $a \ge b$ built into $a^2=b^2+c^2$ , which forces $a^2$ to be the larger of the two ellipse denominators every time.

A hyperbola has no such requirement: its relation is $c^2=a^2+b^2$ , with $a$ and $b$ unconstrained relative to each other, so $a^2$ can sit under either the larger OR the smaller denominator.

$\frac{x^2}{361}-\frac{y^2}{400}=1 \;\Longrightarrow\; x^2 \text{ positive, though } 361<400 \;\Longrightarrow\; \text{opens left and right.}$

What fixes $a^2$ 's position is which term carries a PLUS sign, since $a$ is always measured along the transverse axis, the axis the curve actually crosses. The classmate's rule, borrowed from the wrong curve, gives the wrong answer.

In one line

$a=19$ , $b=20$ , $c=\sqrt{761}$ ; vertices $(\pm19,0)$ , foci $(\pm\sqrt{761},0)$ , asymptotes $y=\pm\frac{20}{19}x$ . The branches open left and right because $x^2$ is the positive term, even though its denominator $361$ is smaller than $400$ : a hyperbola's orientation follows the sign of the term, not the size of the denominator, unlike an ellipse.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Reads $a$ and $b$ correctly from the denominators. . Worth 1 point.

Computes $c$ using $c^2=a^2+b^2$ , not the ellipse's relation, leaving it as an unsimplified radical. . Worth 1 point.

Notes that $c$ comes out largest of the three, the opposite of an ellipse where $a$ is largest. . Worth 1 point.

Part B 3 points

Places the vertices and foci correctly on the transverse axis. . Worth 1 point.

Writes both asymptote equations with the correct slope $\pm\frac{b}{a}$ . . Worth 2 points.

Part C 4 points

States that a hyperbola's orientation follows the positive term, not the larger denominator, and explains that the ellipse's larger-denominator rule follows from a size constraint ( $a \ge b$ ) the hyperbola does not share. . Worth 3 points. needs an explanation, not just an answer

Applies the correct test to this equation, confirming it opens left and right despite $400>361$ . . Worth 1 point.
5. A general equation, a location test, and a question about what scaling changes . 10 points. Question 5 of 10.

A shape is given by $4x^2+4y^2-8x+40y+4=0$ .
1. Part A.
  Classify this equation and convert it to standard form.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 3 points
2. Part B.
  Using your reduced general-form expression from part A (the one with leading coefficient $1$ ), substitute the point $(2,3)$ and decide whether it is inside, on, or outside the circle.
  
  Carry your own answer forward Use your own reduced general form from part A's working, even if a coefficient differs from the one printed here.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
3. Part C.
  Explain why multiplying every term of a general-form equation by a positive constant (as the original equation does, with $4$ ) never changes which points are classified as inside, on, or outside, even though it changes every coefficient $D$ , $E$ , and $F$ .
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

It is a circle: $(x-1)^2+(y+5)^2=25$ .

Part B

$40$ , positive, so $(2,3)$ is outside the circle.

Part C

Multiplying the whole expression by a positive constant multiplies its value at every point by that same positive constant, which cannot change the value's sign, and the inside/on/outside classification depends only on that sign.

Worked solution

Part A

The coefficients on $x^2$ and $y^2$ are equal and share a sign ( $4$ and $4$ ), which is the circle case. Divide by $4$ and complete the square.

$x^2+y^2-2x+10y+1=0 \;\Longrightarrow\; (x-1)^2+(y+5)^2=-1+1+25=25.$

Part B

$2^2+3^2-2(2)+10(3)+1=4+9-4+30+1=40.$

Positive, so the point lies outside the circle.

Part C

Let $g(x,y)=x^2+y^2+Dx+Ey+F$ be the reduced expression and $k>0$ a positive constant. The scaled equation is $k \cdot g(x,y)=0$ , and at any point $(x,y)$ its left side is exactly $k \cdot g(x,y)$ .

$\operatorname{sign}\bigl(k \cdot g(x,y)\bigr) = \operatorname{sign}\bigl(g(x,y)\bigr) \quad \text{for every } k>0.$

Multiplying a real number by a positive constant can change its size but never its sign. Since the location test reads only the SIGN of the expression at a point, scaling the whole equation by a positive constant leaves every classification unchanged, even though $D$ , $E$ , and $F$ themselves are multiplied by $k$ and look different.

In one line

$4x^2+4y^2-8x+40y+4=0$ is a circle, $(x-1)^2+(y+5)^2=25$ ; the point $(2,3)$ gives $40$ in the reduced general form, so it lies outside. Scaling a general-form equation by a positive constant multiplies the expression's value at every point by that same constant, which cannot flip its sign, so the inside/on/outside classification never changes even though $D$ , $E$ , $F$ do.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the equal, same-sign squared-term coefficients as the circle case. . Worth 1 point.

Divides by the leading coefficient and completes the square correctly to reach standard form. . Worth 2 points.

Part B 3 points

Substitutes $(2,3)$ into the reduced general-form expression correctly. . Worth 2 points.

Reads the sign correctly to reach a location verdict. . Worth 1 point.

Part C 4 points

States that multiplying by a positive constant scales the expression's value without changing its sign, for every point. . Worth 3 points. needs an explanation, not just an answer

Connects that fact to the location test depending only on the sign, so the classification is invariant under positive scaling. . Worth 1 point.
6. A general equation and two orientation tests . 13 points. Question 6 of 10.

A shape is given by $16x^2-9y^2-64x-54y-161=0$ .
1. Part A.
  Classify this equation, convert it to standard form, and state its center.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 4 points
2. Part B.
  Find $a$ , $b$ , $c$ , and give the vertices and foci.
  
  Carry your own answer forward Read these off your own standard form and center from part A.
  
  Explain what it means Words, not just symbols. Say what the number is telling you about the situation. 4 points
3. Part C.
  State the asymptote equations. Then compare the two rules for a hyperbola's orientation, checking which denominator is larger versus checking which term is positive: using your own values from part B, decide whether the two rules agree or disagree for THIS equation. Describe, without solving one, what a different pair of denominators would need to look like for the two rules to AGREE.
  
  Carry your own answer forward Use your own $a$ , $b$ , and center from parts A and B.
  
  Compare the two methods Say what each one costs you, and when you would reach for it. 5 points
The answer

Part A

It is a hyperbola: $\dfrac{(x-2)^2}{9}-\dfrac{(y+3)^2}{16}=1$ , center $(2,-3)$ .

Part B

$a=3$ , $b=4$ , $c=5$ . Vertices $(5,-3)$ and $(-1,-3)$ ; foci $(7,-3)$ and $(-3,-3)$ .

Part C

$y+3=\pm\frac{4}{3}(x-2)$ . The rules DISAGREE: the larger denominator ( $16$ ) sits under the negative term, so that rule wrongly says vertical. They agree when the larger denominator sits under the positive term, e.g. $\frac{x^2}{25}-\frac{y^2}{9}=1$ .

Worked solution

Part A

The coefficients on $x^2$ and $y^2$ , $16$ and $-9$ , have opposite signs, which is the hyperbola case. Group, factor, and complete the square.

$16(x^2-4x)-9(y^2+6y)=161.$

Add $4$ inside the first bracket (contributing $16\cdot4=64$ ) and $9$ inside the second (contributing $-9\cdot9=-81$ ).

$16(x-2)^2-9(y+3)^2=161+64-81=144 \;\Longrightarrow\; \frac{(x-2)^2}{9}-\frac{(y+3)^2}{16}=1.$

Center $(2,-3)$ .

Part B

The positive term is $x$ , so $a^2=9$ , $b^2=16$ : $a=3$ , $b=4$ .

$c^2=a^2+b^2=9+16=25 \;\Longrightarrow\; c=5.$

Vertices and foci lie on the horizontal transverse axis, $a$ and $c$ from the center $(2,-3)$ : vertices $(2\pm3,-3)$ , foci $(2\pm5,-3)$ .

Part C

The asymptotes pass through the center with slope $\pm\frac{b}{a}$ (positive $x$ term):

$y-(-3)=\pm\frac{4}{3}(x-2) \;\Longrightarrow\; y+3=\pm\frac{4}{3}(x-2).$

The two orientation rules DISAGREE in this problem: the larger denominator, $16$ , sits under $y^2$ , the term that is NEGATIVE, so the larger-denominator rule would (wrongly) call this hyperbola vertical, while the correct rule, which term is positive, calls it horizontal, and horizontal is what it actually is. The two rules agree only when the larger denominator happens to sit under the positive term. For instance, $\frac{x^2}{25}-\frac{y^2}{9}=1$ has $x^2$ positive AND its denominator, $25$ , is the larger one, so both rules call it horizontal and land on the same answer. No new equation needs to be solved to see this: it only requires choosing denominators where the larger one sits under the positive sign.

In one line

$16x^2-9y^2-64x-54y-161=0$ is a hyperbola, $\frac{(x-2)^2}{9}-\frac{(y+3)^2}{16}=1$ , center $(2,-3)$ , $a=3$ , $b=4$ , $c=5$ , vertices $(5,-3)$ and $(-1,-3)$ , foci $(7,-3)$ and $(-3,-3)$ , asymptotes $y+3=\pm\frac{4}{3}(x-2)$ . The larger-denominator rule and the positive-term rule DISAGREE here, since $16$ sits under the negative term $y^2$ ; a hyperbola like $\frac{x^2}{25}-\frac{y^2}{9}=1$ , where the larger denominator carries the positive sign, shows the two rules agreeing instead, and the positive-term rule is always the correct one.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Identifies the opposite-sign squared-term coefficients as the hyperbola case. . Worth 1 point.

Completes both squares correctly, applying the negative bracket coefficient to what is added to the right side. . Worth 2 points.

States the center correctly. . Worth 1 point.

Part B 4 points

Finds $a$ , $b$ , $c$ correctly using $c^2=a^2+b^2$ . . Worth 2 points.

Shifts the vertices and foci correctly from the center found in part A. . Worth 2 points.

Part C 5 points

Writes both asymptote equations through the correct center with the correct slope. . Worth 2 points.

Explains that the two rules disagree here because the larger denominator sits under the negative term, and constructs (or clearly describes) a case where the larger denominator sits under the positive term instead, so the rules agree. . Worth 3 points. needs an explanation, not just an answer
7. A branch that climbs forever without ever crossing its own guide . 10 points. Question 7 of 10.

A hyperbola has equation $\dfrac{y^2}{400}-\dfrac{x^2}{225}=1$ .
1. Part A.
  Find $a$ , $b$ , $c$ , and the eccentricity.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  State the equations of the two asymptotes.
  
  Carry your own answer forward Use your own $a$ and $b$ from part A.
  
  Write the expression An equation or an expression is enough here. Show how you built it. 2 points
3. Part C.
  Solve $\dfrac{y^2}{400}-\dfrac{x^2}{225}=1$ for $y$ in terms of $x$ (the upper branch), and show that for every $x \ge 0$ , this branch lies strictly above the positive-slope asymptote you found in part B, without ever touching it.
  
  Carry your own answer forward Use your own value of $a^2/b^2$ from part A if it differs from the one printed here.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 5 points
The answer

Part A

$a=20$ , $b=15$ , $c=25$ , $e=\dfrac{5}{4}$ .

Part B

$y=\pm\dfrac{4}{3}x$ .

Part C

$y=\sqrt{400+\tfrac{16}{9}x^2}$ , which is strictly greater than $\sqrt{\tfrac{16}{9}x^2}=\tfrac{4}{3}x$ for every $x \ge 0$ , since adding the positive constant $400$ under a square root strictly increases its value.

Worked solution

Part A

The $y^2$ term is positive: $a^2=400$ , $b^2=225$ , so $a=20$ , $b=15$ .

$c^2=400+225=625 \;\Longrightarrow\; c=25, \qquad e=\frac{c}{a}=\frac{25}{20}=\frac{5}{4}.$

Part B

With $y^2$ positive, the slope is $\pm\frac{a}{b}$ .

$y=\pm\frac{20}{15}x=\pm\frac{4}{3}x.$

Part C

Solve for $y$ on the upper branch.

$\frac{y^2}{400}=1+\frac{x^2}{225} \;\Longrightarrow\; y^2=400+\frac{400}{225}x^2=400+\frac{16}{9}x^2 \;\Longrightarrow\; y=\sqrt{400+\frac{16}{9}x^2}$

(taking the positive root, since this is the upper branch).

Compare to the asymptote: for $x\ge0$ , $\frac{4}{3}x=\sqrt{\left(\frac{4}{3}x\right)^2}=\sqrt{\frac{16}{9}x^2}$ . Since $400>0$ ,

$400+\frac{16}{9}x^2 \;>\; \frac{16}{9}x^2 \;\Longrightarrow\; \sqrt{400+\frac{16}{9}x^2} \;>\; \sqrt{\frac{16}{9}x^2},$

so $y > \frac{4}{3}x$ for every $x\ge0$ . The gap $y-\frac{4}{3}x$ is always strictly positive, however small it becomes for large $x$ , so the branch approaches the asymptote but never reaches it.

In one line

$a=20$ , $b=15$ , $c=25$ , $e=\frac{5}{4}$ , asymptotes $y=\pm\frac{4}{3}x$ . Solving the upper branch gives $y=\sqrt{400+\frac{16}{9}x^2}$ , which strictly exceeds $\frac{4}{3}x=\sqrt{\frac{16}{9}x^2}$ for every $x \ge 0$ , since adding the positive constant $400$ under a square root always increases it: the branch climbs alongside its asymptote without ever touching it.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Finds $a$ and $b$ correctly from the denominators. . Worth 1 point.

Computes $c$ with $c^2=a^2+b^2$ . . Worth 1 point.

Reduces the eccentricity to lowest terms and reports it as a single value. . Worth 1 point.

Part B 2 points

Uses $\pm\frac{a}{b}$ , the correct ratio for this orientation, and reduces it fully. . Worth 2 points.

Part C 5 points

Solves for $y$ correctly on the upper branch, keeping the positive square root. . Worth 2 points.

Compares the two square-root expressions and argues, from the positive constant $400$ under the root, that the branch's $y$ -value is strictly greater than the asymptote's for every $x \ge 0$ . . Worth 3 points. needs an explanation, not just an answer
8. A line and an ellipse . 11 points. Question 8 of 10.

An ellipse has equation $\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$ .
1. Part A.
  Find $a$ , $b$ , $c$ , and give the vertices and foci.
  
  Solve and show your work Write each step out, and end with the value and its units. 3 points
2. Part B.
  Substitute the line $y=x-5$ into the ellipse's equation, collect the resulting quadratic in $x$ , and use its discriminant to state how many points the line shares with the ellipse.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain, in general, what the SIGN of a quadratic's discriminant determines about the number of real intersection points, and what its SQUARE-STATUS (whether it is a perfect square) determines separately about the coordinates of those points. Check both against your own discriminant from part B, and state what property the coefficients need for your argument about square-status to make sense.
  
  Carry your own answer forward Use your own discriminant value from part B, whatever it came out to be.
  
  Explain why it works A sentence or two. Reasons, not steps. 4 points
The answer

Part A

$a=6$ , $b=2\sqrt{5}$ , $c=4$ . Vertices $(\pm6,0)$ , foci $(\pm4,0)$ .

Part B

$14x^2-90x+45=0$ ; discriminant $5580$ , positive, so the line meets the ellipse at two points.

Part C

The SIGN of $b^2-4ac$ alone decides whether the roots are real and distinct, real and repeated, or complex, before any square root is taken. Its SQUARE-STATUS is a separate question, deciding only whether real roots are rational or irrational, and that second question needs $a$ , $b$ , $c$ themselves to be rational.

Worked solution

Part A

The larger denominator, $36$ , sits under $x^2$ : $a^2=36$ , $b^2=20$ , so $a=6$ , $b=\sqrt{20}=2\sqrt5$ .

$c^2=36-20=16 \;\Longrightarrow\; c=4.$

Vertices $(\pm6,0)$ , foci $(\pm4,0)$ .

Part B

Multiply by $180$ , the least common multiple of $36$ and $20$ , to clear both fractions at once.

$5x^2+9(x-5)^2=180 \;\Longrightarrow\; 5x^2+9x^2-90x+225=180 \;\Longrightarrow\; 14x^2-90x+45=0.$

$(-90)^2-4(14)(45)=8100-2520=5580>0.$

Positive, so the line is a secant, meeting the ellipse at two points.

Part C

The quadratic formula's classification of roots, two real, one repeated, or a complex pair, rests entirely on the SIGN of $b^2-4ac$ , decided before any square root is ever taken. A positive value of any size guarantees two distinct real roots by that sign test alone.

$5580>0 \;\Longrightarrow\; \text{two distinct real roots}.$

Whether that positive value is a perfect square is a separate question, controlling only whether $\sqrt{b^2-4ac}$ itself comes out rational (giving rational roots) or irrational (giving roots that are real but not expressible as simple fractions). This second question needs $a$ , $b$ , $c$ to be rational numbers in the first place, since "rational versus irrational" is not even the right pair of options for roots built from irrational coefficients.

$5580 \text{ is a positive integer, not a perfect square} \;\Longrightarrow\; \sqrt{5580} \notin \mathbb{Q} \;\Longrightarrow\; \text{the two roots are irrational.}$

But the two intersection points are still exactly two, and still exactly real, because that verdict was already settled by the sign of $5580$ alone, before its square root was ever considered. For any quadratic with RATIONAL coefficients, however messy, the sign test and the perfect-square test remain two different questions.

In one line

$a=6$ , $b=2\sqrt{5}$ , $c=4$ , vertices $(\pm6,0)$ , foci $(\pm4,0)$ . The line $y=x-5$ substituted into the ellipse gives $14x^2-90x+45=0$ , discriminant $5580>0$ , so the line is a secant meeting the ellipse at two points. That the discriminant is a positive, non-square integer only means the two points have irrational coordinates; the sign of the discriminant alone, decided before any square root, already guarantees they are real and distinct.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the larger denominator as $a^2$ and simplifies $b=\sqrt{20}$ to $2\sqrt{5}$ . . Worth 1 point.

Computes $c$ correctly by subtraction. . Worth 1 point.

Reports the vertices and foci on the correct axis, matching the orientation found. . Worth 1 point.

Part B 4 points

Substitutes the line correctly and collects the result into a single quadratic in $x$ . . Worth 2 points.

Computes the discriminant correctly. . Worth 1 point.

States the resulting count and type (two points, a secant) from the sign of the discriminant. . Worth 1 point.

Part C 4 points

Separates the SIGN test (which decides real versus complex, and distinct versus repeated) from the PERFECT-SQUARE test (which decides rational versus irrational roots), rather than treating them as one question. . Worth 3 points. needs an explanation, not just an answer

States that the perfect-square test only makes sense for quadratics with rational coefficients, and that with that restriction the separation holds for any such coefficients, not just this particular equation. . Worth 1 point.
9. A dish, a strut, and the definition doing the checking . 11 points. Question 9 of 10.

A satellite dish is shaped like a parabola opening upward, with its vertex at $(0,0)$ on a cross-sectional diagram measured in centimeters. Its focal length (vertex-to-focus distance) is $18$ cm.
1. Part A.
  Write the equation of the dish's cross-section.
  
  Model the situation Name your unknown first, then write every other quantity in terms of that one letter. 3 points
2. Part B.
  A support strut attaches to the dish at the point where $x=12$ cm from the vertex line. How far above the vertex does the strut attach, and how far is that point from the focus?
  
  Carry your own answer forward Use your own equation from part A.
  
  Solve and show your work Write each step out, and end with the value and its units. 4 points
3. Part C.
  Explain, using the definition of a parabola (not the shortcut formula), why the strut's distance to the focus found in part B must equal its distance to the directrix, and state the directrix's equation.
  
  Carry your own answer forward Use your own $p$ and the strut's height from parts A and B.
  
  Justify your claim State the claim, then give the reason it has to be true. 4 points
The answer

Part A

$x^2=72y$ .

Part B

$y=2$ cm above the vertex; focal distance $20$ cm.

Part C

By definition, every point of a parabola is equidistant from the focus and the directrix, so the strut's focal distance, $20$ cm, must equal its distance to the directrix $y=-18$ ; checking directly, the strut sits at height $y=2$ , which is $2-(-18)=20$ cm above that line, confirming it.

Worked solution

Part A

The vertex is at the origin and the focal length is $p=18$ , with the parabola opening upward, so it takes the form $x^2=4py$ .

$x^2=4(18)y=72y.$

Part B

Substitute $x=12$ .

$12^2=72y \;\Longrightarrow\; 144=72y \;\Longrightarrow\; y=2.$

The focus is $(0,18)$ . The shortcut $PF=y+p$ gives

$PF=2+18=20 \text{ cm},$

and the distance formula agrees: $\sqrt{(12-0)^2+(2-18)^2}=\sqrt{144+256}=\sqrt{400}=20$ cm.

Part C

A parabola is DEFINED as the set of points equidistant from a focus and a directrix; the shortcut $PF=y+p$ is not an extra fact layered on top of that definition, it is exactly what the distance to the directrix simplifies to for a point above a directrix at $y=-p$ .

The directrix here is $y=-p=-18$ . The strut sits at $(12,2)$ , so its distance to the directrix is

$2-(-18)=20 \text{ cm},$

exactly matching the focal distance of $20$ cm found in part B by the shortcut. The two distances agree not by coincidence but because the shortcut IS the directrix-distance computation, restated, and the defining condition of the parabola sets that equal to the focal distance for every point of the curve.

In one line

$x^2=72y$ ; the strut attaches at $(12,2)$ , $2$ cm above the vertex, and $20$ cm from the focus. Its distance to the directrix $y=-18$ is also $2-(-18)=20$ cm, matching exactly, because the shortcut $PF=y+p$ is the directrix-distance side of the parabola's own defining condition, not a separate fact.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 3 points

Identifies the focal length as $p$ and writes the coefficient as $4p$ , not $p$ alone. . Worth 2 points.

Reports the correct equation $x^2=72y$ . . Worth 1 point.

Part B 4 points

Substitutes $x=12$ correctly and solves for $y=2$ . . Worth 2 points.

Finds the focal distance correctly, whether by the shortcut $PF=y+p$ or by the distance formula directly. . Worth 1 point.

Reports both the height and the focal distance with the cm unit. . Worth 1 point.

Part C 4 points

States the parabola's defining condition (equidistant from focus and directrix) and identifies that the shortcut is exactly the directrix-distance side of that condition. . Worth 2 points. needs an explanation, not just an answer

Computes the directrix $y=-18$ and verifies the strut's distance to it directly, confirming it matches part B. . Worth 2 points.
10. Two relations that point opposite ways . 12 points. Question 10 of 10.

An ellipse and a hyperbola obey opposite relations: $c^2=a^2-b^2$ for the ellipse, $c^2=a^2+b^2$ for the hyperbola.
1. Part A.
  Consider a general ellipse centered at the origin with foci $(\pm c,0)$ and semi-minor axis $b$ , so its co-vertex is $(0,b)$ . Using only the fact that the co-vertex is equidistant from both foci, and that the sum of its two focal distances is $2a$ , show that each focal distance from the co-vertex equals $a$ exactly, and hence that $a^2=b^2+c^2$ .
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 4 points
2. Part B.
  Now take as GIVEN that a general hyperbola centered at the origin with foci $(\pm c,0)$ , vertices $(\pm a,0)$ , and conjugate semi-axis $b$ satisfies $c^2=a^2+b^2$ (established, as in this chapter, from the hyperbola's own difference-of-distances condition). Using that given relation, verify that the central rectangle's corner $(a,b)$ (the rectangle reaching $a$ along the transverse axis and $b$ along the conjugate axis) sits exactly a distance $c$ from the center.
  
  Complete the derivation Each line should follow from the one above it. Say what lets you take each step. 3 points
3. Part C.
  Part A derived $a^2=b^2+c^2$ from a genuine POINT ON THE ELLIPSE, the co-vertex, using the curve's own sum condition. Part B, by contrast, had to take $c^2=a^2+b^2$ as GIVEN before verifying a fact about the rectangle's corner. Explain specifically why the corner cannot be used the way Part A used the co-vertex, that is, why no argument starting from a curve point's own defining condition was available to derive the relation from the corner directly.
  
  Carry your own answer forward Refer to your own proofs from parts A and B.
  
  Justify your claim State the claim, then give the reason it has to be true. 5 points
The answer

Part A

Since the co-vertex $(0,b)$ is equidistant from both foci and its two focal distances sum to $2a$ , each distance is $a$ ; computing that distance directly gives $\sqrt{b^2+c^2}=a$ , so $a^2=b^2+c^2$ .

Part B

The distance from the origin to the corner $(a,b)$ is $\sqrt{a^2+b^2}$ by the Pythagorean theorem, and since $c^2=a^2+b^2$ is given, that distance equals $\sqrt{c^2}=c$ exactly.

Part C

The corner is not a point on the hyperbola, so the curve's own difference condition cannot apply to it as the sum condition applied to the ellipse's co-vertex; the corner is well-defined only once $a$ , $b$ , $c$ are already known, so Part B verifies the relation rather than deriving it.

Worked solution

Part A

The co-vertex $(0,b)$ sits on the $y$ -axis, which is the perpendicular bisector of the segment joining $(\pm c,0)$ , so it is equidistant from both foci. Call that common distance $d$ . Since the SUM of the two focal distances is always $2a$ on an ellipse, and both distances equal $d$ here,

$d+d=2a \;\Longrightarrow\; d=a.$

Now compute $d$ directly, using the distance formula from $(0,b)$ to $(c,0)$ :

$d=\sqrt{(c-0)^2+(0-b)^2}=\sqrt{c^2+b^2}.$

Setting the two expressions for $d$ equal:

$\sqrt{b^2+c^2}=a \;\Longrightarrow\; a^2=b^2+c^2.$

Part B

The distance from the center (the origin) to the corner $(a,b)$ is, by the Pythagorean theorem,

$\sqrt{(a-0)^2+(b-0)^2}=\sqrt{a^2+b^2}.$

Since $c^2=a^2+b^2$ is given, $\sqrt{a^2+b^2}=\sqrt{c^2}=c$ . So the corner-to-center distance is exactly $c$ : the central rectangle's half-diagonal matches the focal distance, which is why the rectangle is a useful picture for the hyperbola's three lengths, not a source of the relation itself.

Part C

Part A's argument works because the co-vertex is a genuine point ON THE ELLIPSE, so the curve's own defining sum condition applies to it directly, the same way it applies to every other point of the curve.

$\text{co-vertex on curve} \;\Longrightarrow\; PF_1+PF_2=2a \text{ applies to it directly.}$

The central rectangle's corner in part B is not a point on the hyperbola at all. It is an auxiliary construction whose coordinates ARE $a$ and $b$ , so asking a curve point's defining condition (the difference $|PF_1-PF_2|=2a$ ) to say something about it does not even make sense: the corner has no pair of focal distances to take a difference of, because it never claims to lie on the curve. The only way to know the corner sits a distance $c$ from the center is to already know $c^2=a^2+b^2$ , which is exactly what Part B assumed as given.

That is why $c^2=a^2+b^2$ genuinely needs to be established the way the ellipse's relation was in Part A: starting from the hyperbola's own difference condition at a real point of the curve (for instance a vertex, or a general point, followed by clearing the two radicals by squaring twice), the way the lesson derives the standard-form equation. Part B's rectangle argument, once that relation is available, is a useful geometric PICTURE of it, not a second independent route to it.

In one line

For an ellipse, the co-vertex is equidistant from both foci and its two focal distances sum to $2a$ , so each equals $a$ ; computing that distance directly gives $a^2=b^2+c^2$ , i.e. $c^2=a^2-b^2$ , a genuine derivation from a curve point. For a hyperbola, GIVEN $c^2=a^2+b^2$ , the central rectangle's corner $(a,b)$ checks out as exactly $c$ from the center by the Pythagorean theorem, but this is a verification, not a derivation: the corner is not a point on the hyperbola, so the curve's own difference condition cannot be applied to it the way the sum condition was applied to the ellipse's co-vertex. The relation itself is established the same way the ellipse's was, from the difference condition at a genuine curve point.

Rubric

Mark your own paper against this. Give yourself the points for each element you actually wrote down, not the ones you meant to.

Part A 4 points

Argues that the co-vertex is equidistant from both foci (by the perpendicular-bisector symmetry) and uses the sum condition to conclude each distance equals $a$ . . Worth 2 points. needs an explanation, not just an answer

Computes the co-vertex-to-focus distance directly and equates it to $a$ to reach $a^2=b^2+c^2$ . . Worth 2 points.

Part B 3 points

Computes the center-to-corner distance as $\sqrt{a^2+b^2}$ using the Pythagorean theorem on the rectangle's half-width $a$ and half-height $b$ . . Worth 2 points.

Uses the GIVEN relation $c^2=a^2+b^2$ , rather than re-deriving it, to identify that distance as exactly $c$ . . Worth 1 point. needs an explanation, not just an answer

Part C 5 points

Identifies that the central rectangle's corner is not a point on the hyperbola, so the curve's own difference condition cannot be applied to it the way the sum condition applied to the ellipse's genuine curve point (the co-vertex). . Worth 3 points. needs an explanation, not just an answer

States that Part B's argument is a verification of the already-given relation, not an independent derivation of it, and correctly identifies what an independent derivation would need (the difference condition at a genuine curve point). . Worth 2 points. needs an explanation, not just an answer

Marking rubrics

Work every question first. Then mark your own paper against these, giving yourself the points for each element you actually wrote down.

1. One equation, read two ways from two given lines 12 points

Part A 4

Writes the distance to the focus and the distance to the directrix as expressions in $x$ and $y$ , and sets them equal before squaring. 2
Squares both sides, expands correctly, and simplifies to isolate $(y-4)^2$ , showing the cancellation of the $x^2$ terms. 2

Part B 4

Reads the vertex $(1,4)$ and computes $p=-4$ correctly from $4p=-16$ , not from $-16$ directly. 2
States the parabola opens left, tying that to the sign of $p$ . 1
Recovers the focus and directrix and confirms they match the stem. 1

Part C 4

Redoes the derivation with general letters ( $u=x-h$ , focus $h+p$ , directrix $h-p$ ) rather than appealing only to the specific numbers above. 2
Identifies that the coefficient $4p$ comes from the difference of squares $(u+p)^2-(u-p)^2=4up$ , and states that this holds for every choice of $p$ and $h$ . 2

2. A range, a general form, and a shortcut worth questioning 12 points

Part A 4

Writes the distance condition $(x-2)^2+(y+5)^2=49$ with correct signs on each coordinate. 2
Expands both squares correctly and collects every term onto one side to reach general form. 2

Part B 4

Substitutes $(6,-2)$ into the general-form expression from part A correctly and computes the resulting value. 2
Reads the sign correctly to reach a location verdict. 2

Part C 4

Identifies that the general-form expression equals the squared distance from the center minus $r^2$ , so its sign carries the same information as the direct distance comparison. 3
Concludes explicitly that neither method skips real work; they perform the same comparison in a different order. 1

3. Two ellipses, read cold, and compared without a formula 10 points

Part A 3

Identifies the larger denominator as $a^2$ and the major axis as running along that variable. 1
Computes $c$ by subtraction, and reports the vertices and foci on the correct axis. 2

Part B 3

Identifies the larger denominator as $a^2$ and the major axis as running along that variable, independently of part A's orientation. 1
Computes $c$ correctly and simplifies $b=\sqrt{40}$ to $2\sqrt{10}$ , and reports the vertices and foci on the correct axis. 2

Part C 4

Identifies elongation with the RATIO $b/a$ (close to $1$ near-circular, close to $0$ flattened) rather than the raw difference $a-b$ , and explains why a ratio, not a difference, is the right measure since it does not change under uniform scaling. 2
Computes the two specific ratios (about $0.385$ versus about $0.904$ ) and reaches the correct verdict that the first ellipse is more elongated. 2

4. A hyperbola, read plainly, and a rule tested against its own exception 10 points

Part A 3

Reads $a$ and $b$ correctly from the denominators. 1
Computes $c$ using $c^2=a^2+b^2$ , not the ellipse's relation, leaving it as an unsimplified radical. 1
Notes that $c$ comes out largest of the three, the opposite of an ellipse where $a$ is largest. 1

Part B 3

Places the vertices and foci correctly on the transverse axis. 1
Writes both asymptote equations with the correct slope $\pm\frac{b}{a}$ . 2

Part C 4

States that a hyperbola's orientation follows the positive term, not the larger denominator, and explains that the ellipse's larger-denominator rule follows from a size constraint ( $a \ge b$ ) the hyperbola does not share. 3
Applies the correct test to this equation, confirming it opens left and right despite $400>361$ . 1

5. A general equation, a location test, and a question about what scaling changes 10 points

Part A 3

Identifies the equal, same-sign squared-term coefficients as the circle case. 1
Divides by the leading coefficient and completes the square correctly to reach standard form. 2

Part B 3

Substitutes $(2,3)$ into the reduced general-form expression correctly. 2
Reads the sign correctly to reach a location verdict. 1

Part C 4

States that multiplying by a positive constant scales the expression's value without changing its sign, for every point. 3
Connects that fact to the location test depending only on the sign, so the classification is invariant under positive scaling. 1

6. A general equation and two orientation tests 13 points

Part A 4

Identifies the opposite-sign squared-term coefficients as the hyperbola case. 1
Completes both squares correctly, applying the negative bracket coefficient to what is added to the right side. 2
States the center correctly. 1

Part B 4

Finds $a$ , $b$ , $c$ correctly using $c^2=a^2+b^2$ . 2
Shifts the vertices and foci correctly from the center found in part A. 2

Part C 5

Writes both asymptote equations through the correct center with the correct slope. 2
Explains that the two rules disagree here because the larger denominator sits under the negative term, and constructs (or clearly describes) a case where the larger denominator sits under the positive term instead, so the rules agree. 3

7. A branch that climbs forever without ever crossing its own guide 10 points

Part A 3

Finds $a$ and $b$ correctly from the denominators. 1
Computes $c$ with $c^2=a^2+b^2$ . 1
Reduces the eccentricity to lowest terms and reports it as a single value. 1

Part B 2

Uses $\pm\frac{a}{b}$ , the correct ratio for this orientation, and reduces it fully. 2

Part C 5

Solves for $y$ correctly on the upper branch, keeping the positive square root. 2
Compares the two square-root expressions and argues, from the positive constant $400$ under the root, that the branch's $y$ -value is strictly greater than the asymptote's for every $x \ge 0$ . 3

8. A line and an ellipse 11 points

Part A 3

Identifies the larger denominator as $a^2$ and simplifies $b=\sqrt{20}$ to $2\sqrt{5}$ . 1
Computes $c$ correctly by subtraction. 1
Reports the vertices and foci on the correct axis, matching the orientation found. 1

Part B 4

Substitutes the line correctly and collects the result into a single quadratic in $x$ . 2
Computes the discriminant correctly. 1
States the resulting count and type (two points, a secant) from the sign of the discriminant. 1

Part C 4

Separates the SIGN test (which decides real versus complex, and distinct versus repeated) from the PERFECT-SQUARE test (which decides rational versus irrational roots), rather than treating them as one question. 3
States that the perfect-square test only makes sense for quadratics with rational coefficients, and that with that restriction the separation holds for any such coefficients, not just this particular equation. 1

9. A dish, a strut, and the definition doing the checking 11 points

Part A 3

Identifies the focal length as $p$ and writes the coefficient as $4p$ , not $p$ alone. 2
Reports the correct equation $x^2=72y$ . 1

Part B 4

Substitutes $x=12$ correctly and solves for $y=2$ . 2
Finds the focal distance correctly, whether by the shortcut $PF=y+p$ or by the distance formula directly. 1
Reports both the height and the focal distance with the cm unit. 1

Part C 4

States the parabola's defining condition (equidistant from focus and directrix) and identifies that the shortcut is exactly the directrix-distance side of that condition. 2
Computes the directrix $y=-18$ and verifies the strut's distance to it directly, confirming it matches part B. 2

10. Two relations that point opposite ways 12 points

Part A 4

Argues that the co-vertex is equidistant from both foci (by the perpendicular-bisector symmetry) and uses the sum condition to conclude each distance equals $a$ . 2
Computes the co-vertex-to-focus distance directly and equates it to $a$ to reach $a^2=b^2+c^2$ . 2

Part B 3

Computes the center-to-corner distance as $\sqrt{a^2+b^2}$ using the Pythagorean theorem on the rectangle's half-width $a$ and half-height $b$ . 2
Uses the GIVEN relation $c^2=a^2+b^2$ , rather than re-deriving it, to identify that distance as exactly $c$ . 1

Part C 5

Identifies that the central rectangle's corner is not a point on the hyperbola, so the curve's own difference condition cannot be applied to it the way the sum condition applied to the ellipse's genuine curve point (the co-vertex). 3
States that Part B's argument is a verification of the already-given relation, not an independent derivation of it, and correctly identifies what an independent derivation would need (the difference condition at a genuine curve point). 2

Multiple choice

Free response