Conic Sections: Chapter Test

20 multiple-choice questions and 10 core practice problems, drawn from across the chapter and mixed together.

Multiple choice

20 questions, 100 points in total, 5 points each. Answer in any order and change your mind as often as you like. When you submit, your answers lock and every question shows its worked solution.

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Question 1 of 20

1

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A parabola with vertex at the origin has equation $y^2 = 40x$ . What is $p$ ?

Answer choices for question 1 A $40$ B $20$ C $10$ D $160$
2

A circle has center $(-4, 2)$ and radius $6$ . Which equation is its standard form?

Answer choices for question 2 A $(x+4)^2+(y-2)^2=36$ B $(x-4)^2+(y+2)^2=36$ C $(x+4)^2+(y-2)^2=6$ D $(x-4)^2+(y+2)^2=6$
3

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $\dfrac{x^2}{36}+\dfrac{y^2}{100}=1$ , which axis is major, and what are $a$ , $b$ , and $c$ ?

Answer choices for question 3 A major axis horizontal; $a=6$ , $b=10$ , $c=8$ B major axis vertical; $a=10$ , $b=6$ , $c=\sqrt{136}$ C major axis vertical; $a=10$ , $b=8$ , $c=6$ D major axis vertical; $a=10$ , $b=6$ , $c=8$
4

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $\dfrac{y^2}{25}-\dfrac{x^2}{49}=1$ , which way do the branches open, and what are $a$ and $b$ ?

Answer choices for question 4 A opens left and right, since $49>25$ ; $a=7$ , $b=5$ B opens up and down; $a=5$ , $b=7$ C opens up and down; $a=7$ , $b=5$ D opens left and right; $a=5$ , $b=7$
5

Which type of conic does $3x^2+3y^2-12x+18y-9=0$ describe, before any further simplification?

Answer choices for question 5 A circle B ellipse C parabola D hyperbola
6

Advanced. This question goes beyond core Algebra II. It is not required by the course.

The equation $y^2-6y-8x+41=0$ describes a parabola. After completing the square, what is $p$ ?

Answer choices for question 6 A $8$ B $-8$ C $4$ D $2$
7

Advanced. This question goes beyond core Algebra II. It is not required by the course.

An ellipse has $a=41$ and $c=40$ . Find its eccentricity, and state whether this ellipse is closer to a circle or to a flattened sliver.

Answer choices for question 7 A $e=\frac{41}{40}$ , which is impossible for an ellipse, so no such curve exists B $e=\frac{40}{41}\approx0.98$ , close to a flattened sliver C $e=\frac{40}{41}\approx0.98$ , close to a circle D $e=\frac{9}{41}\approx0.22$ , close to a circle
8

A circle has general form $x^2+y^2-10x+2y+13=0$ . What does substituting the point $(1,-2)$ into the left side tell you about its location?

Answer choices for question 8 A $4$ , so the point lies inside the circle B $-4$ , so the point lies inside the circle C $4$ , so the point lies outside the circle D $0$ , so the point lies on the circle
9

Substituting the line $y=2x+10$ into $x^2+y^2+2x-6y-15=0$ and simplifying gives $x^2+6x+5=0$ . How many points does the line share with the circle?

Answer choices for question 9 A $0$ , the line misses the circle B $1$ , a tangent C infinitely many D $2$ , a secant
10

Advanced. This question goes beyond core Algebra II. It is not required by the course.

An ellipse has foci $(\pm 9,0)$ , and the sum of the focal distances from any point on it is $30$ . What is its standard equation?

Answer choices for question 10 A $\dfrac{x^2}{225}+\dfrac{y^2}{144}=1$ B $\dfrac{x^2}{144}+\dfrac{y^2}{225}=1$ C $\dfrac{x^2}{225}+\dfrac{y^2}{81}=1$ D $\dfrac{x^2}{900}+\dfrac{y^2}{819}=1$
11

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A hyperbola has $a=9$ and $b=12$ . What is $c$ ?

Answer choices for question 11 A $\sqrt{63}$ B $21$ C $15$ D $225$
12

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A parabola has focus $(3,5)$ and directrix $y=-1$ . What is its standard equation?

Answer choices for question 12 A $(x-3)^2=3(y-2)$ B $(x-3)^2=12(y-2)$ C $(x-3)^2=12(y-5)$ D $(y-2)^2=12(x-3)$
13

Completing the square on $2x^2+2y^2+20x-8y+2=0$ gives which standard form?

Answer choices for question 13 A $(x+10)^2+(y-4)^2=114$ B $(x+5)^2+(y-2)^2=28$ C $(x+5)^2+(y-2)^2=30$ D $(x-5)^2+(y+2)^2=28$
14

Advanced. This question goes beyond core Algebra II. It is not required by the course.

What are the asymptote slopes of $\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$ ?

Answer choices for question 14 A $\pm\dfrac{4}{3}$ B $\pm\dfrac{9}{16}$ C $\pm\dfrac{16}{9}$ D $\pm\dfrac{3}{4}$
15

Substituting the line $y=x+1$ into $\dfrac{x^2}{4}-\dfrac{y^2}{9}=1$ and clearing fractions gives $5x^2-8x-40=0$ . How many points does the line share with the hyperbola?

Answer choices for question 15 A $2$ , a secant B $0$ , the line misses the hyperbola C $1$ , a tangent D $2$ points algebraically, but only one actually lies on the curve
16

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A parabola has equation $(x+2)^2=-20(y-6)$ . A classmate reports its directrix as the point $(-2,11)$ . What is the directrix actually?

Answer choices for question 16 A the point $(-2,11)$ , as the classmate said B $y=1$ C $y=11$ D $x=-2$
17

Advanced. This question goes beyond core Algebra II. It is not required by the course.

A hyperbola has $a=16$ and $c=65$ , with $x^2$ the positive term. Find its eccentricity and its asymptote slope.

Answer choices for question 17 A $e=\dfrac{65}{16}$ , slope $\dfrac{63}{16}$ B $e=\dfrac{16}{65}$ , slope $\dfrac{63}{16}$ C $e=\dfrac{65}{16}$ , slope $\dfrac{16}{63}$ D $e=\dfrac{65}{63}$ , slope $\dfrac{63}{16}$
18

Which type of conic does $5x^2-3y^2+20x+6y-2=0$ describe?

Answer choices for question 18 A ellipse B hyperbola C parabola D circle
19

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $\dfrac{x^2}{20}+\dfrac{y^2}{45}=1$ , a classmate reports the foci as $(\pm5,0)$ . Where are the foci actually?

Answer choices for question 19 A $(\pm5,0)$ , as the classmate said B $(0,\pm\sqrt{65})$ C $(0,\pm25)$ D $(0,\pm5)$
20

For what positive value of $k$ is the line $y=x+k$ tangent to $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ ?

Answer choices for question 20 A $k=25$ B $k=7$ C $k=5$ D $k=\sqrt{7}$

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Where you stand

State the focus-directrix distance condition that defines a parabola, and use it to find $p$ , the focus, and the directrix from a standard-form equation, distinguishing $p$ from the coefficient $4p$ . Review: Parabolas
Derive a circle's equation from the distance condition $PC = r$ , convert between standard form and general form $x^2+y^2+Dx+Ey+F=0$ , and use the sign of the general-form expression at a point to decide whether that point is inside, on, or outside the circle. Review: Circles
Read $a$ , $b$ , $c$ , and the orientation of the major axis from an ellipse's standard equation, using $c^2 = a^2 - b^2$ and the rule that $a^2$ is always the larger denominator. Review: Ellipses
Read $a$ , $b$ , $c$ , and the orientation of a hyperbola from its standard equation, using $c^2 = a^2 + b^2$ and the rule that $a^2$ sits under the positive term regardless of which denominator is larger, and find the asymptote slopes. Review: Hyperbolas
Classify a general second-degree equation as a circle, an ellipse, a hyperbola, or a parabola from the coefficients of its squared terms, and convert it to standard form by completing the square. Review: Circles , Ellipses , Hyperbolas , Parabolas
Compute the eccentricity $e = \frac{c}{a}$ and use it to place a conic on the single scale that sorts a circle, an ellipse, a parabola, and a hyperbola. Review: Ellipses , Hyperbolas
Substitute a line into a conic's equation, collect the resulting quadratic, and use its discriminant to decide whether the line is a secant, a tangent, or misses the curve. Review: Circles
Build the equation of a conic from geometric data, such as a focus and directrix, or two foci together with a sum, difference, or vertex distance. Review: Ellipses , Hyperbolas , Parabolas

Core practice

10 problems from across the chapter. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A sideways curve

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $2(y+1)^2=-5(x-4)$ , give $p$ , the focus, and the directrix, and state the distance condition its points satisfy.
1. Hint 1
  The signed vertex-to-focus displacement is encoded in the parabola's standard form.
2. Hint 2
  The focus and directrix lie on opposite sides of the vertex along the horizontal axis.
Answer

$p=-\frac58$ ; focus $(\frac{27}{8},-1)$ ; directrix $x=\frac{37}{8}$ ; equal distances to that focus and line.

Full solution

Dividing by $2$ gives
$(y+1)^2=-\frac52(x-4).$
Thus $4p=-5/2$ , so $p=-5/8$ .
The vertex is $(4,-1)$ ; adding p horizontally gives the focus $(27/8,-1)$ , and subtracting p gives the directrix $x=37/8$ .
Every point has equal distance to this focus and this line.

Answer

$p=-\frac58$ ; focus $(\frac{27}{8},-1)$ ; directrix $x=\frac{37}{8}$ ; equal distances to that focus and line.

Key idea

After the squared factor has coefficient 1, the coefficient of the unsquared factor is 4p, where p, not zero, is the signed vertex-to-focus displacement.
Problem 1: Mark complete
Problem 2 A fixed-distance boundary

A boundary consists of all points 3 units from $(-2,-1)$ . Write its equation in general form with coefficients of $x^2$ and $y^2$ equal to 1. Is $(0,0)$ inside, on, or outside the boundary?
1. Hint 1
  A fixed positive distance from one point defines a circle.
2. Hint 2
  The general-form expression equals squared distance minus squared radius.
Answer

$x^2+y^2+4x+2y-4=0$ ; $(0,0)$ is inside.

Full solution

The distance rule gives
$(x+2)^2+(y+1)^2=9.$
Expanding and moving $9$ left gives
$x^2+y^2+4x+2y-4=0.$
At $(0,0)$ the expression is $-4$ , which is negative, so the point is inside.
Its squared distance is $5$ , less than $9$ .

Answer

$x^2+y^2+4x+2y-4=0$ ; $(0,0)$ is inside.

Key idea

The sign of a normalized circle expression compares squared distance with squared radius.
Problem 2: Mark complete
Problem 3 An oval boundary

Advanced. This question goes beyond core Algebra II. It is not required by the course.

The figure shows an ellipse. Give $a$ , $b$ , and $c$ , and identify the direction of its major axis.

The ellipse.

Text description of this figure

A grid with the x-axis from -5 to 6 and the y-axis from -2 to 5 on equal scales, a grid line and a label at every integer. A single ellipse is drawn: its leftmost point is at x equals -4 and its rightmost at x equals 5, both at height 1.5, and its lowest point is at y equals -1 and its highest at y equals 4, both above x equals 0.5. No center, focus, axis or equation is marked.
1. Hint 1
  Read the distances from the center to the horizontal and vertical extremes.
2. Hint 2
  The larger semiaxis is a, and $c^2=a^2-b^2$ .
Answer

$a=\frac92$ , $b=\frac52$ , $c=\sqrt{14}$ ; major axis horizontal.

Full solution

The center is $(\frac12,\frac32)$ , midway between the extremes.
The horizontal extremes are $\frac92$ units from it and the vertical extremes $\frac52$ units, so $a=\frac92$ and $b=\frac52$ .
The focal relation gives
$c^2=\frac{81}4-\frac{25}4$ $c^2=14,$
so
$c=\sqrt{14}.$
The longer direction is horizontal.

Answer

$a=\frac92$ , $b=\frac52$ , $c=\sqrt{14}$ ; major axis horizontal.

Key idea

Read semiaxes from the center before using the ellipse focal relation.
Problem 3: Mark complete
Problem 4 A two-branch equation

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $\frac{(x+3)^2}{4}-\frac{(y-2)^2}{5}=1$ , give $a$ , $b$ , $c$ , the opening direction, and both asymptote slopes.
1. Hint 1
  The positive term determines both the opening and the location of a squared.
2. Hint 2
  Use $c^2=a^2+b^2$ ; the rectangle's vertical half-size divided by its horizontal half-size gives the asymptote slope magnitude.
Answer

$a=2$ , $b=\sqrt5$ , $c=3$ ; opens left and right; slopes $\pm\frac{\sqrt5}{2}$ .

Full solution

The positive x term gives $a^2=4$ , and the negative term gives $b^2=5$ .
Therefore
$c^2=4+5$ $c^2=9.$
So $a=2$ , $b=\sqrt5$ , and $c=3$ .
The branches open horizontally, and the slopes are
$\pm b/a=\pm\sqrt5/2.$

Answer

$a=2$ , $b=\sqrt5$ , $c=3$ ; opens left and right; slopes $\pm\frac{\sqrt5}{2}$ .

Key idea

In standard form with right side 1 and positive denominators, a squared is the denominator of the positive squared term, whichever denominator is larger.
Problem 4: Mark complete
Problem 5 A general equation

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Classify $2x^2+7y^2-28x+70y+245=0$ . Give its standard form, center, and foci.
1. Hint 1
  Equal signs on unequal squared coefficients suggest an ellipse; completing the squares verifies it.
2. Hint 2
  Normalize to a right side of 1 before finding the major axis and focal distance.
Answer

Ellipse; $\frac{(x-7)^2}{14}+\frac{(y+5)^2}{4}=1$ ; center $(7,-5)$ ; foci $(7\pm\sqrt{10},-5)$ .

Full solution

Group the terms and complete both squares: $2(x-7)^2-98+7(y+5)^2-175=-245$ , so
$2(x-7)^2+7(y+5)^2=28,$
since
$-245+98+175=28.$
Dividing by $28$ gives
$\frac{(x-7)^2}{14}+\frac{(y+5)^2}{4}=1.$
This is an ellipse centered at $(7,-5)$ , with horizontal major axis.
Since $c^2=14-4$ , which is $10$ , the foci are $(7\pm\sqrt{10},-5)$ .

Answer

Ellipse; $\frac{(x-7)^2}{14}+\frac{(y+5)^2}{4}=1$ ; center $(7,-5)$ ; foci $(7\pm\sqrt{10},-5)$ .

Key idea

Completing and normalizing both squares exposes a conic's geometry.
Problem 5: Mark complete
Problem 6 A quadratic boundary

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Classify $7y^2-9x^2-84y-90x-36=0$ . Give its standard form, center, and asymptote equations.
1. Hint 1
  Opposite signs on the squared terms suggest a hyperbola, with a degenerate case to check.
2. Hint 2
  Keep the negative coefficient attached while completing the x square.
Answer

Hyperbola; $\frac{(y-6)^2}{9}-\frac{(x+5)^2}{7}=1$ ; center $(-5,6)$ ; $y-6=\pm\frac{3}{\sqrt7}(x+5)$ .

Full solution

Completing the squares, $7(y-6)^2-252-9(x+5)^2+225=36$ , so
$7(y-6)^2-9(x+5)^2=63,$
since
$36+252-225=63.$
The positive right side is nonzero.
Dividing by $63$ gives
$\frac{(y-6)^2}{9}-\frac{(x+5)^2}{7}=1.$
This hyperbola is centered at $(-5,6)$ and opens vertically.
Its rectangle has vertical half-size $3$ and horizontal half-size $\sqrt7$ , so the asymptote slopes are $\pm\frac{3}{\sqrt7}$ , giving the stated lines.

Answer

Hyperbola; $\frac{(y-6)^2}{9}-\frac{(x+5)^2}{7}=1$ ; center $(-5,6)$ ; $y-6=\pm\frac{3}{\sqrt7}(x+5)$ .

Key idea

Opposite squared-term signs and a nonzero normalized constant produce a hyperbola.
Problem 6: Mark complete
Problem 7 A path and a boundary

A circle consists of points $\sqrt8$ units from the origin. A straight path has coordinates $x=1+t$ and $y=3-t$ , where t is real. Give the circle's standard equation and decide whether the path is a secant, a tangent, or misses the circle. Give any shared points.
1. Hint 1
  A shared point must satisfy both the circle's distance condition and the path equations.
2. Hint 2
  The discriminant of the resulting quadratic in t counts the intersections.
Answer

$x^2+y^2=8$ ; tangent at $(2,2)$ .

Full solution

The distance rule gives
$x^2+y^2=8.$
Substituting the path yields
$(1+t)^2+(3-t)^2=8.$
This becomes $2t^2-4t+2=0$ , whose discriminant is
$\Delta=16-4(2)(2)$ $\Delta=0.$
The repeated solution is $t=1$ , giving $(2,2)$ .
Its squared distance is $8$ , so the path is tangent there.

Answer

$x^2+y^2=8$ ; tangent at $(2,2)$ .

Key idea

A repeated real root after line substitution gives a single tangency point on a circle.
Problem 7: Mark complete
Problem 8 The geometric specification

Advanced. This question goes beyond core Algebra II. It is not required by the course.

The figure gives the focus and directrix of a parabola. Find its standard equation, p, and vertex. State the equality of distances satisfied by a general point on the curve.

The focus F and the directrix.

Text description of this figure

A grid with the x-axis from -5 to 4 and the y-axis from -1 to 7 on equal scales, a grid line and a label at every integer, and an extra tick on the x-axis labeled one half. A single point labeled F sits at x equals -2 and y equals 3. A dashed vertical line runs the full height of the grid through that one-half tick, at x equals one half, and is labeled directrix. No parabola, vertex or distance is drawn.
1. Hint 1
  Read the focus and line, then locate their midpoint along the perpendicular axis.
2. Hint 2
  Use the signed displacement from the vertex to the focus for p.
Answer

$(y-3)^2=-5(x+\frac34)$ ; $p=-\frac54$ ; vertex $(-\frac34,3)$ ; $\sqrt{(x+2)^2+(y-3)^2}=|x-\frac12|$ .

Full solution

The focus is $(-2,3)$ and the directrix is $x=\frac12$ , so their midpoint along the axis is the vertex $(-\frac34,3)$ .
The signed displacement from the vertex to the focus is $p=-\frac54$ , so $4p=-5$ and the standard equation is
$(y-3)^2=-5(x+\tfrac34).$
The distance equality is
$\sqrt{(x+2)^2+(y-3)^2}=|x-\frac12|.$
Squaring it cancels the matching x-squared terms and gives the same standard equation.

Answer

$(y-3)^2=-5(x+\frac34)$ ; $p=-\frac54$ ; vertex $(-\frac34,3)$ ; $\sqrt{(x+2)^2+(y-3)^2}=|x-\frac12|$ .

Key idea

For a parabola the focus lies off the directrix, the vertex bisects the perpendicular segment between them, and the curve opens toward the focus.
Problem 8: Mark complete
Problem 9 Two related curves

Advanced. This question goes beyond core Algebra II. It is not required by the course.

Curve E is $\frac{x^2}{3}+\frac{y^2}{11}=1$ . Curve H is $\frac{x^2}{8}-\frac{y^2}{6}=1$ . A student argues that both vertices of H lie inside E because H has the smaller value of $a$ . Is this correct? Identify E's major-axis direction, H's transverse-axis direction, and H's vertices.
1. Hint 1
  The larger ellipse denominator and the positive hyperbola term decide the two directions.
2. Hint 2
  Test a vertex of H in the left side of E's equation.
Answer

Incorrect; E's major axis is vertical, H's transverse axis is horizontal, and H's vertices $(\pm2\sqrt2,0)$ both lie outside E.

Full solution

E's larger denominator, $11$ , is under the y term, so its major axis is vertical, with
$a=\sqrt{11}.$
H's positive term is the x term, so it opens left and right, with $a=\sqrt8=2\sqrt2$ and vertices $(\pm2\sqrt2,0)$ .

At a vertex of H, the left side of E's equation is $\frac83$ , which is greater than $1$ , so both vertices lie outside E.
H's value of $a$ is smaller, but it runs along E's minor axis, where E reaches only $\sqrt3$ .

Answer

Incorrect; E's major axis is vertical, H's transverse axis is horizontal, and H's vertices $(\pm2\sqrt2,0)$ both lie outside E.

Key idea

Comparing two conics' sizes needs the directions of their axes, not only the values of a.
Problem 9: Mark complete
Problem 10 Eccentricity with a parameter

Advanced. This question goes beyond core Algebra II. It is not required by the course.

For $t>0$ , consider $\frac{y^2}{5}-\frac{x^2}{t}=1$ . Express its eccentricity in terms of $t$ . Can it be less than $1$ ? Find $t$ when $e=\sqrt{\frac{14}5}$ .
1. Hint 1
  For a hyperbola, the denominator under the positive term gives a squared.
2. Hint 2
  Find c squared from the sum of the denominators, then compare e squared with 1.
Answer

$e=\sqrt{1+\frac t5}$ ; no, it exceeds $1$ for every $t>0$ ; $t=9$ .

Full solution

The positive y term gives $a^2=5$ , and $c^2=5+t$ .
So
$e^2=\frac{5+t}5,$
that is,
$e=\sqrt{1+\frac t5}.$

Since $t>0$ , $e^2>1$ and so $e>1$ : a hyperbola's eccentricity is never less than $1$ .
Setting $1+\frac t5=\frac{14}5$ gives $\frac t5=\frac95$ , so $t=9$ .

Answer

$e=\sqrt{1+\frac t5}$ ; no, it exceeds $1$ for every $t>0$ ; $t=9$ .

Key idea

A hyperbola's eccentricity exceeds one because its focal distance c is larger than its semitransverse length a.
Problem 10: Mark complete

Star problems Ten optional challenges to stretch your reasoning Advanced. This problem set goes beyond core Algebra II. You can skip it. → Continue to Chapter 13: Probability and Statistics →

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