Question

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

Answer 1

Given that eccentricity i.e, e=3/2 and `(pmae,0)=(pm2,0)`
`therefore ae=2`
`rArr acdot3/2=2rArra=4//3`
`becauseb^(2)=a^(2)(e^(2)-1)`
`rArr b^(2)=16/9(9/4-1)`
`rArrb^(2)=16/4(5/4)=+20/9`
So, the hyperbola is
`(x^(2))/(16/9)-(y^(2))/(20/9)=1`
`rArr (x^(2))/4-(y^(2))/5=4/9`