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Hyperbola Worksheet

Hyperbola Worksheet
  • Page 1
 1.  
Find the eccentricity of the hyperbola 9y² - 16x² = 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.


Correct answer : (2)
 2.  
Find the vertices of the hyperbola 16x2 - y2 = 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).


Correct answer : (4)
 3.  
Find an equation of the hyperbola with foci (- 5, 0) and (5, 0) whose conjugate axis has length 6.
a.
x²16 - y²9 = 1
b.
16x² - y² = 16
c.
x²9 - y²16 = 1
d.
x² - 16y² = 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


Correct answer : (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.
(x - 4)²9 - (y - 4)²16 = 1
c.
x² - 16y² = 16
d.
(y - 4)²16 - (x - 4)²9 = 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.


Correct answer : (4)
 5.  
Find the foci of the hyperbola (x - 4)²256 - (y + 9)²144 = 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).


Correct answer : (2)
 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.


Correct answer : (3)
 7.  
Find an equation in the standard form for the hyperbola with center at (0, 0), a = 4, e = 5 4 and with horizontal focal axis.
a.
16x² - y² = 16
b.
9x² - y² = 9
c.
x²9 - y²16 = 1
d.
x²16 - y²9 = 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.


Correct answer : (4)
 8.  
Find the vertices of the hyperbola 25x² - 16y² - 50x - 32y - 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).


Correct answer : (2)
 9.  
What are the foci of the hyperbola 81x2 - 144y2 = 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).]


Correct answer : (3)
 10.  
What are the x - intercepts of the hyperbola x²16 - y²16 = 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).


Correct answer : (2)

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