The distance between the directrices of the hyperbola `x=8s e ctheta,\ y=8\ t a ntheta,` a.`8sqrt(2)` b. `16sqrt(2)` c. `4sqrt(2)` d. `6sqrt(2)`

Explanation

We have given `x=8sectheta,y=8tantheta`

The eccentricity of the hyperbola

`x^2/a^2−y^2/b^2=1` is given by,

`e=sqrt(1+b^2/a^2)`

and the distance between the directrices is given by
`(2a)/e`.

Here, `a=8` and `b=8` since `sec^2theta−tan^2theta=1`

Therefore, the distance becomes

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