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Equation of the hyperbola with eccentric...

Equation of the hyperbola with eccentricity `3/2` and foci at `(±2,0)` is

A

`(x^(2))/4-(y^(2))/5=4/9`

B

`(x^(2))/4-(y^(2))/9=4/9`

C

`(x^(2))/4-(y^(2))/9=1`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
Given,that eccentricity of the hyperbola e=3/2
and foci(`pm`2,0),(`pm`ae,0)
`because ae=2`
`rArr axx3//2=2rArra=4//3`
`because b^(2)=a^(2)(e^(2)-1)`
`rArrb^(2)=16/9(9/4-1)rArrb^(2)=16/9(5/4)`
`rArrb^(2)=20/9`
So, the equation of the hyperbola is
`(x^(2))/(16/9)-(y^(2))/(20//9)=1rArr(x^(2))/4-(y^(2))/5=4/9`
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