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Find the equation of hyperbola : whose a...

Find the equation of hyperbola : whose axes are coordinate axes and the distances of one of its vertices from the foci are 3 and 1

A

`3x^(2) -y^(2) =3`

B

`x^(2)-3y^(2) +3 =0`

C

`x^(2)-3y^(2) -3 =0`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A, B
Consider `(x^(2))/(a^(2)) - (y^(2))/(b^(2)) =1`
Let one of the vertices be `(a,0)`
Foci are `(+- ae,0)`
According to the question we have
`ae - a = 1` and `ae + a=3`
Solving we get `a = 1` and `e =2`
`:. e^(2) =1 + (b^(2))/(a^(2)) rArr b^(2) =3`
`:.` Equation of hyperbola is `(x^(2))/(1) -(y^(2))/(3) =1` or `3x^(2) - y^(2) =3`
For `(x^(2))/(a^(2)) -(y^(2))/(b^(2)) =-1` is `e^(2) =1 + (a^(2))/(b^(2)) rArr b^(2) = (1)/(3)`
`:.` Hyperbola will be `x^(2) - 3y^(2) + 3=0`
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