Find the equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13.

Options

A. `25x^(2)-144y^(2)=900`
B. `-25x^(2)+144y^(2)=900`
C. `144x^(2)-25y^(2)=900`
D. `-144x^(2)+25y^(2)=900`

Answer

A

Explanation

Let `2a` and `2b` be the transverse and conjugate axes and `e` be the eccentricity. Let the centre be the origin and the transverse and the conjugate axes the coordinate axes.
Then, the equation of the hyperbola is
`(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`....`(i)`
We have, `2b=5` and `2ae=13`.
`:.b^(2)=a^(2)(e^(2)-1)`
`impliesb^(2)=a^(2)e^(2)-a^(2)implies(25)/(4)=(169)/(4)-a^(2)impliesa^(2)=(144)/(4)impliesa=6`
Substituting the values of `a` and b` in `(i)`, the equation of the hyperbola is
`(x^(2))/(36)-(y^(2))/(25//4)=1` or , `25x^(2)-144y^(2)=900`