1.

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

Answer»

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


Discussion

No Comment Found

Related InterviewSolutions