Question

Write the distance between the directrices of the hyperbola.

$x=8sec\theta ,y=8tan\theta .$

Answer 1

$Givenx=8sec\theta ,y=8tan\theta .$

$sec\theta =\frac{x}{8}andtan\theta =\frac{y}{8}$

$Now,se{c}^2\theta -ta{n}^2\theta =1$

$\Rightarrow {\left(\frac{x}{8}\right)}^2-{\left(\frac{y}{8}\right)}^2$

$\Rightarrow \frac{x^2}{64}-\frac{y^2}{64}=1...\left(1\right)$

$\text{On comparing (1) with equation of hyperbola i.e}\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,weget$

$a=8andb=8$

$\text{Now,eccentricity can be find out as}$

$b^2=a^2(e^2-1)$

$8^2=8^2(e^2-1)$

$1=e^2-1$

$2=e^2$ $e=\sqrt{2}$

$\text{Now,distance between their directrix}$

$=\frac{a}{e}-(-\frac{a}{e})=\frac{a}{e}+\frac{a}{e}=2\frac{a}{e}$

$=2\left(\frac{8}{\sqrt{2}}\right)=8\sqrt{2}$