Core practice ← Back to lesson

Hyperbolas: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 The rectangle and branches

The figure shows a hyperbola and its central rectangle. Find the two asymptote equations.

The hyperbola and its central rectangle.

Text description of this figure

A grid with the x-axis from -6 to 4 and the y-axis from -8 to 10 on equal scales, a grid line at every integer and a label at every even number. A hyperbola opens up and down: its upper branch turns at y equals 5 and its lower branch at y equals -3, both directly above or below x equals -1, and the branches spread outward as they leave the window. A dashed rectangle has its left and right sides at x equals -3 and x equals 1 and its bottom and top at y equals -3 and y equals 5, touching the two turning points; its center is marked C at x equals -1 and y equals 1. No diagonal, asymptote or equation is drawn.
1. Hint 1
  The asymptotes extend the diagonals of the central rectangle.
2. Hint 2
  Read the center and calculate the rise divided by the run from the center to a corner.
Answer

$y-1=\pm2(x+1)$ .

Full solution

The rectangle center is $(-1,1)$ .
Its half-width is $2$ and its half-height is $4$ , so the diagonal slopes are
$m=\pm\frac42$ $m=\pm2.$
Both lines pass through $(-1,1)$ , giving
$y-1=\pm2(x+1).$

Answer

$y-1=\pm2(x+1)$ .

Key idea

The diagonals of a hyperbola's central rectangle determine its asymptotes.
Problem 1: Mark complete
Problem 2 Two measured distances

At one point on a hyperbola, the distances to its foci are 23 units and 10 units. Find the length of its transverse axis.
1. Hint 1
  The constant absolute difference of focal distances is the full transverse-axis length.
2. Hint 2
  Subtract the shorter distance from the longer distance.
Answer

13 units.

Full solution

The absolute difference is
$|23-10|=13.$
This is the constant $2a$ , which is also the length of the transverse axis, so the required length is $13$ units.

Answer

13 units.

Key idea

A hyperbola's transverse-axis length equals its constant absolute focal-distance difference.
Problem 2: Mark complete
Problem 3 A ratio in the record

A hyperbola has $b/a=\sqrt3$ , where $a$ and $b$ are positive. Find its eccentricity.
1. Hint 1
  Eccentricity compares the distances from the center to a focus and to a vertex.
2. Hint 2
  Divide $c^2=a^2+b^2$ by $a^2$ , use $e=c/a$ , and take the positive square root.
Answer

$e=2$ .

Full solution

Dividing the focal relation by $a^2>0$ gives
$e^2=1+\left(\frac ba\right)^2.$
Hence
$e^2=1+3$ $e^2=4.$
Eccentricity is positive, so $e=2$ , which is greater than $1$ as required.

Answer

$e=2$ .

Key idea

A hyperbola's eccentricity satisfies e squared equals one plus the square of the whole ratio b over a.
Problem 3: Mark complete
Problem 4 An outline to sketch

Sketch $\frac{(y+1)^2}{9}-\frac{(x-2)^2}{4}=1$ using its central rectangle. Give the rectangle's corners, the vertices, and the asymptote equations.
1. Hint 1
  The positive y term determines the opening, while the center shifts every construction point.
2. Hint 2
  The vertical half-size is a and the horizontal half-size is b; extend the rectangle's diagonals.
Answer

Corners $(0,-4),(4,-4),(0,2),(4,2)$ ; vertices $(2,-4),(2,2)$ ; asymptotes $y+1=\pm\frac32(x-2)$ .

Full solution

The center is $(2,-1)$ , with $a=3$ vertically and $b=2$ horizontally.
Moving by those half-sizes gives the four corners in the answer.
The positive y term puts the vertices $3$ units above and below the center.
Each diagonal has slope
$m=\pm\frac32,$
so its equation is
$y+1=\pm\frac32(x-2).$
The completed sketch shows branches opening up and down from the vertices, approaching these lines.

The hyperbola, its central rectangle and its asymptotes.

Answer

Corners $(0,-4),(4,-4),(0,2),(4,2)$ ; vertices $(2,-4),(2,2)$ ; asymptotes $y+1=\pm\frac32(x-2)$ .

Key idea

A vertical hyperbola uses a as its rectangle's vertical half-size and b as its horizontal half-size.
Problem 4: Mark complete
Problem 5 A missing denominator

The hyperbola $\frac{x^2}{4}-\frac{y^2}{q}=1$ , with $q>0$ , has foci 8 units apart. Find $q$ and the absolute difference of focal distances for its points.
1. Hint 1
  Half the focal separation is c, while the positive term gives a.
2. Hint 2
  Use $c^2=a^2+b^2$ to recover the denominator q.
Answer

$q=12$ ; absolute difference 4 units.

Full solution

The focal separation gives $c=4$ , and the positive term gives $a=2$ .
Therefore
$q=4^2-2^2$ $q=12.$
The defining absolute difference is $2a=4$ units.
Checking, $a^2+b^2=4+12$ , which is $16$ , that is, $c^2$ .

Answer

$q=12$ ; absolute difference 4 units.

Key idea

For a hyperbola, b squared equals c squared minus a squared, with a squared read under the positive term of the standard form.
Problem 5: Mark complete
Problem 6 A location from two totals

A point P lies on the right branch of $\frac{x^2}{25}-\frac{y^2}{24}=1$ . Its distances to the two foci add to 28 units. Find all possible coordinates of P.
1. Hint 1
  The hyperbola also fixes the difference of the two focal distances.
2. Hint 2
  Find both distances, then subtract their squared coordinate expressions to isolate x.
Answer

$(10,6\sqrt2)$ and $(10,-6\sqrt2)$ .

Full solution

Here $a=5$ and $c^2=25+24=49$ , so $c=7$ .
On the right branch the left-focus distance exceeds the right-focus distance by $10$ .
A sum of $28$ therefore gives distances $19$ and $9$ .

The foci are $(-7,0)$ and $(7,0)$ .
Subtracting the squared distances, $(x+7)^2-(x-7)^2$ equals $361-81$ , so
$28x=280$ $x=10.$
Substitute into the hyperbola:
$4-\frac{y^2}{24}=1.$
Thus $y^2=72$ , giving
$y=\pm6\sqrt2.$
The squared focal distances are $289+72$ and $9+72$ , that is, $361$ and $81$ , as required.

Answer

$(10,6\sqrt2)$ and $(10,-6\sqrt2)$ .

Key idea

Combining a focal sum with a hyperbola's fixed difference can determine a point on either side of its axis.
Problem 6: Mark complete
Problem 7 A parameter in the equation

For real $t$ , consider $x^2+8x=2y^2+4y-t$ . Find the center when the graph is a hyperbola. Determine which values of $t$ give a horizontal hyperbola, a vertical hyperbola, or two intersecting lines.
1. Hint 1
  Complete the square in each variable, keeping the factor 2 on the y group.
2. Hint 2
  Inspect the sign of the completed equation's right side, including zero.
Answer

Center $(-4,-1)$ ; horizontal for $t<14$ , vertical for $t>14$ , and the two lines $x+4=\pm\sqrt2(y+1)$ for $t=14$ .

Full solution

Completing the squares, $x^2+8x=(x+4)^2-16$ and $2y^2+4y=2(y+1)^2-2$ , so
$(x+4)^2-2(y+1)^2=14-t.$
The center is $(-4,-1)$ .

If $14-t>0$ , dividing by it leaves the x term positive, so the hyperbola opens left and right: $t<14$ .
If $14-t<0$ , dividing by it makes the y term positive, so it opens up and down: $t>14$ .
If $t=14$ , then $(x+4)^2=2(y+1)^2$ , the two lines
$x+4=\pm\sqrt2(y+1).$

Answer

Center $(-4,-1)$ ; horizontal for $t<14$ , vertical for $t>14$ , and the two lines $x+4=\pm\sqrt2(y+1)$ for $t=14$ .

Key idea

The sign of the completed equation's right side decides a hyperbola's opening, and zero gives a pair of lines.
Problem 7: Mark complete
Problem 8 Two geometry records

An ellipse has $a=6$ and $c=4$ . A hyperbola has $a=4$ and $c=6$ . A student says both curves have the same value of $b$ . Is this correct? Explain why the subtraction is different for the two curves.
1. Hint 1
  Identify the largest length in the right-triangle relation for each curve.
2. Hint 2
  For the ellipse a is the hypotenuse; for the hyperbola c is the hypotenuse.
Answer

Correct; both have $b=2\sqrt5$ .

Full solution

The ellipse uses $b^2=a^2-c^2$ , giving
$b^2=36-16$ $b^2=20.$
The hyperbola uses $b^2=c^2-a^2$ , also giving $b^2=20$ .
Thus both have $b=\sqrt{20}$ , which is $2\sqrt5$ .
The same numerical subtraction occurs because a and c exchanged roles, not because the two curve types use the same relation.

Answer

Correct; both have $b=2\sqrt5$ .

Key idea

Ellipse and hyperbola focal relations place different letters in the hypotenuse role.
Problem 8: Mark complete
Problem 9 Lines parallel to an asymptote

For real $t$ , how many points do $\frac{x^2}{49}-\frac{y^2}{4}=1$ and $y=\frac27x+t$ share? Explain what changes when $t=0$ .
1. Hint 1
  An exact common point would have to satisfy both equations.
2. Hint 2
  After substituting, collect the terms in x; handle a zero coefficient separately.
Answer

None when $t=0$ , where the line is an asymptote; exactly one for every $t\ne0$ .

Full solution

Substituting gives
$\frac{x^2}{49}-\frac{(2x/7+t)^2}{4}=1.$
The $x^2$ terms cancel, leaving $-\frac{tx}7-\frac{t^2}4=1$ , that is,
$4tx=-28-7t^2.$

If $t=0$ , this reads $0=-28$ , so there is no common point: the line is the asymptote
$y=\frac27x.$
If $t\ne0$ , it gives exactly one x, $x=-\frac{28+7t^2}{4t}$ , and so exactly one shared point.

Answer

None when $t=0$ , where the line is an asymptote; exactly one for every $t\ne0$ .

Key idea

A line parallel to an asymptote meets the hyperbola exactly once, except the asymptote itself, which never meets it.
Problem 9: Mark complete
Problem 10 The information in two lines

A student claims the asymptotes $y=3x$ and $y=-3x$ uniquely determine a hyperbola centered at the origin and opening left and right. Decide whether the claim is correct. Give two different standard equations if it is false.
1. Hint 1
  Which equation links the asymptote slopes to a and b?
2. Hint 2
  Test what happens to that equation when both lengths are doubled.
Answer

False; for example, $x^2-\frac{y^2}{9}=1$ and $\frac{x^2}{4}-\frac{y^2}{36}=1$ .

Full solution

The required slopes fix $b/a=3$ .
Taking $a=1$ , $b=3$ gives
$x^2-\frac{y^2}{9}=1.$
Taking $a=2$ , $b=6$ gives
$\frac{x^2}{4}-\frac{y^2}{36}=1.$
Both open horizontally and have slopes $\pm3$ , but their vertices are $(\pm1,0)$ and $(\pm2,0)$ , so they are different hyperbolas.

Answer

False; for example, $x^2-\frac{y^2}{9}=1$ and $\frac{x^2}{4}-\frac{y^2}{36}=1$ .

Key idea

For a fixed opening direction, the asymptotes determine the center and the ratio b over a, while the size remains free.
Problem 10: Mark complete

← Back to lesson