﻿ Hyperbola Worksheet | Problems & Solutions

# Hyperbola Worksheet

Hyperbola Worksheet
• Page 1
1.
Find the eccentricity of the hyperbola 9$y$² - 16$x$² = 144.
 a. 1.6 b. 1.25 c. 1.23 d. 1.56

#### Solution:

9y² - 16x² = 144

y²16 - x²9 = 1.
[Divide both sides by 144.]

This is in the standard form y²a² - x²b² = 1 where a² = 16 and b² = 9 and hence a = 4, b = 3.

The eccentricity of the hyperbola = e = (a² + b²)a
[Use the formula for eccentricity.]

= (16 + 9)4 = 5 / 4 = 1.25.

2.
Find the vertices of the hyperbola 16$x$2 - $y$2 = 16.
 a. (± 4, 0) b. (0, ± 1) c. (0, ± 4) d. (± 1, 0 )

#### Solution:

16x2 - y2 = 16

x21 - y216 = 1
[Divide both sides by 16.]

This is in the standard form x2a2 - y2b2 = 1 where a2 = 1, b2 = 16

a = 1, b = 4
[Solve for a, b.]

So, the vertices of the hyperbola are (± a, 0 ) = ( ± 1, 0).

3.
Find an equation of the hyperbola with foci (- 5, 0) and (5, 0) whose conjugate axis has length 6.
 a. $\frac{x²}{16}$ - $\frac{y²}{9}$ = 1 b. 16$x$² - $y$² = 16 c. $\frac{x²}{9}$ - $\frac{y²}{16}$ = 1 d. $x$² - 16$y$² = 16

#### Solution:

Foci of the hyperbola are F1 = (- 5, 0) and F2 = (5, 0)

Center of the hyperbola = Midpoint of F1F2 = [(- 5 + 5)2, (0 + 0)2] = (0, 0).
[Use the mid point formula.]

Distance between foci = 2c = F1F2 = (5 + 5)² + 0² = 10 and hence c = 5.

So, the foci are on the x - axis with c = 5.

Semi conjugate axis is b = 6 / 2 = 3.

a² = c² - b² = 25 - 9 = 16.
[Pythagorean relation.]

So, the standard form of the equation for the hyperbola is x²16 - y²9 = 1

4.
Find the standard form of the equation for the hyperbola whose transverse axis has end points (4, 8) and (4, 0) and whose conjugate axis has length 6.
 a. ($x$ - 4)² - ($y$ - 4)² = 1 b. - = 1 c. $x$² - 16$y$² = 16 d. - = 1

#### Solution:

End points of the transverse axis are A(4, 8) and B(4, 0).

Center of the hyperbola = Mid point of AB = (h, k) = [(4 + 4)2, (8 + 0)2] = (4, 4).
[Use the mid point formula.]

Slope of AB = (0 - 8)(4 - 4) = - 8 / 0 which is undefined.
[Use slope formula.]

Since slope of transverse axis is undefined, it is a vertical line.

Length of transverse axis = 2a = (4 - 4)² + (8 - 0)² = 8 and hence a = 4.

Length of the conjugate axis = 2b = 6 and hence b = 3.

So, the equation of the hyperbola in the standard form is (y - k)²a² - (x -h)²b² = 1 or (y - 4)²16 - (x - 4)²9 = 1.

5.
Find the foci of the hyperbola - = 1.
 a. (- 24, 9) and (-16, -9) b. (24, -9) and (-16, - 9) c. (24, 9) and (-16, 9) d. (-24, 9) and (-16, 9)

#### Solution:

(x - 4)²256 - (y + 9)²144 = 1.

This is in the standard form (x - h)²a² - (y -  k)²b² = 1 where h = 4, k = -9, a² = 256 and b² = 144

a = 16, b = 12
[Solve for a, b.]

c = (a² + b²) = (256 + 144) = 20.
[Pythagorean relation.]

So, the foci are (h ± c, k) = ( 4 ± 20, - 9) or (24, -9) and (-16, -9).

6.
If $e$ is the eccentricity of a hyperbola, then which of the following is correct?
 a. $e$ = 1 b. 0 ≤ $e$ < 1 c. $e$ > 1 d. $e$ = 0

#### Solution:

The eccentricity of a hyperbola = e = (a² + b²)a where ′a′ is the semi transverse axis, ′b′ is the semi conjugate axis.
[Definition.]

Since (a² + b²) > a, (a² + b²)a > 1 and hence e > 1.

7.
Find an equation in the standard form for the hyperbola with center at (0, 0), $a$ = 4, $e$ = $\frac{5}{4}$ and with horizontal focal axis.
 a. 16$x$² - $y$² = 16 b. 9$x$² - $y$² = 9 c. $\frac{x²}{9}$ - $\frac{y²}{16}$ = 1 d. $\frac{x²}{16}$ - $\frac{y²}{9}$ = 1

#### Solution:

Center of the hyperbola = (0, 0)

Focal axis of the hyperbola is horizontal, a = 4, e = 5 / 4 .

e = (a² + b²)a
[Use the formula for eccentricity.]

5 / 4 = (16 + b²)4
[Substitute values of a, e.]

25 = 16 + b².

b² = 9.

So the equation of the hyperbola in the standard form is x²a² - y²b² = 1 or x²16 - y²9 = 1.

8.
Find the vertices of the hyperbola 25$x$² - 16$y$² - 50$x$ - 32$y$ - 391 = 0.
 a. (5, 1) and (3, 1) b. (5, - 1) and (- 3, - 1) c. (5, 1) and (- 3, 1) d. (- 5, 1) and (3, - 1)

#### Solution:

25x² - 16y² - 50x - 32y - 391 = 0

25(x² - 2x) - 16(y² + 2y) = 391
[Group the terms.]

25(x² - 2x + 1) - 16(y² + 2y + 1) = 391 + 25 - 16
[Complete the squares.]

25(x - 1)² - 16(y + 1)² = 400

(x - 1)²16 - (y + 1)²25 = 1
[Divide both sides by 400.]

This is in the standard form (x - h)²a² - (y - k)²b² = 1 where a² = 16, b² = 25, h = 1, k = - 1

a = 4, b = 5
[Solve for a, b.]

The vertices of the hyperbola = (h ± a, k) = (1 ± 4, - 1) = (5, - 1) and (- 3, - 1).

9.
What are the foci of the hyperbola 81$x$2 - 144$y$2 = 11664?
 a. (± 12, 0) b. (0, ± 15) c. (± 15, 0) d. (± 108, 0)

#### Solution:

81x2 - 144y2 = 11664
[Equation of hyperbola.]

x²144 - y²81 = 1
[Divide both sides by 11664.]

This is in the standard form x²a² - y²b² = 1 of a hyperbola, where a2 = 144, b2 = 81.

c2 = a2 + b2 = 144 + 81 = 225
[Pythagorean relation.]

c = 15
[As c > 0.]

So, the foci of the given hyperbola are (± 15 , 0).
[Use foci : (± c, 0).]

10.
What are the $x$ - intercepts of the hyperbola $\frac{x²}{16}$ - $\frac{y²}{16}$ = 1?
 a. (0, - 16), (0, 16) b. (- 4, 0), (4, 0) c. (0, - 4), (0, 4) d. (- 16, 0), (16, 0)

#### Solution:

x²16 - y²16 = 1
[Equation of hyperbola.]

x²16 - 0 = 1
[Substitute 0 for y for x - intercepts.]

x2 = 16

x = - 4 or x = 4
[Solve for x.]

So, the x - intercepts of the given hyperbola are (- 4 , 0) and (4, 0).