In each of the following find the equations of the hyperbola satisfying the given condition: vertices `(0,+-6),\ e=5/3`

Explanation

We have given foci `(0,+-6)` and `e=5/3`

Therefore, the equation of the hyperbola is of the form

`x^2/a^2− y^2/b^2=1`

Since the foci are `(0,+-6),b=+-6`.

We know that
`a^2=6^2((5/3)^2-1)`

`a^2=36((25/9)-1)`

`a^2=(36(25-9))/9`

`impliesa^2=64`

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