Question

A hyperbola has a vertical transverse axis of length 16 and asymptotes of y = 6 5 x − 9 and y = − 6 5 x − 4 . Find the center of the hyperbola, its focal length, and its eccentricity. The center of the hyperbola is (Incorrect , Incorrect ). The focal length is . The eccentricity is​

Answer 1

Explanation:

A

a=8

a^2=64

..

slope of asymptotes=8/5=b/a

b=8a/5=64/5

b^2=(64/5)^2=163.84

..

c^2=a^2+b^2=64+163.84=227.84

c=√227.84≈15.09 (focal length)

..

eccentricity:c/a=15.09/8≈1.89

..

Finding center:

Equations of asymptotes are straight lines that intersect at the center.

Solve as system of two equations:

y=8x/5+2

y=-8x/5+2

..

5y=8x+10

5y=-8x+10

add

10y=20

y=2

..

subtract

0=16x+0

x=0

..

Center: (0,2)

focal length: 15.09

eccentricity:1.89

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