Question

If the normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at any point $P(a\cos \theta ,b\tan \theta )$ meets the transverse and conjugate axes in $G$ and $g$ respectively and if ${}^{\prime }{f}^{\prime }$ is the foot of perpendicular to the normal at $P$ from the centre ${}^{\prime }{C}^{\prime }$ then

Read the following carefully and answer the following questions

The length of $P{G}^2$ is

• A. $\frac{b^2}{a^2}(b^{2}co{\sec }^2\theta +a^2{\cot }^2\theta )$
• B. $\frac{a^2}{b^2}(b^{2}co{\sec }^2\theta +a^2{\cot }^2\theta )$
• C. $\frac{b^2}{a^2}(b^2{\sec }^2\theta +a^2{\tan }^2\theta )$
• D. $\frac{b^2}{a^2}(b^2{\tan }^2\theta +a^2{\sec }^2\theta )$

Answer 1

The correct option is C $\frac{b^2}{a^2}(b^2{\sec }^2\theta +a^2{\tan }^2\theta )$

Equation of the tangent is
$\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1$

And normal is $ax\cos \theta +by\cot \theta =a^2+b^2$

Then
normal at $P$ meets the co-ordinate axes at
$G\left(\begin{array}{c}\frac{a^2+b^2}{a}\sec \theta ,0\end{array}\right)\text{and}g\left(\begin{array}{c}0,\frac{a^2+b^2}{a}\tan \theta \end{array}\right)$

$\therefore P{G}^2={\left(\begin{array}{c}\frac{a^2+b^2}{a}\sec \theta -a\sec \theta \end{array}\right)}^2+(b\tan \theta -0{)}^2$

$\Rightarrow P{G}^2=\frac{b^2}{a^2}(b^2{\sec }^2\theta +a^2{\tan }^2\theta )$