Write the distance between the directrices of the hyperbola x=8sec theta and y=8 tan theta

Question

Write the distance between the directrices of the hyperbola x=8secθ and y=8 tanθ?

Answer 1

Given, x=8secθ and y=8tanθ
=> secθ=x/8 and tanθ=y/8
Now, we know that sec²θ-tan²θ=1
=> (x/8)²-(y/8)²=1
=> x²/64-y²/64=1 -------(1)
on comparing (1) with equation of hyperbola i.e.,
x²/a²-y²/b²=1, we get
a=8 and b=8
Now, eccentricity van be find out as
b²=a²(e²-1)
=> 8²=8²(e²-1)
=> 1=e²-1
=> e=√2
Now, distance between their directrix
=a/e-(-a/e)=a/e+a/e=2a/e
= 2(8/√2)=8√2