Question

A normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the axes in $M$ and $N$ and lines $MP$ and $NP$ are drawn perpendiculars to the axes meeting at $P$. Then the locus of $P$ is :

• A. $a^2{x}^2-b^2{y}^2={(a^2-b^2)}^2$
• B. $a^2{x}^2-b^2{y}^2={(a^2+b^2)}^2$
• C. $b^2{y}^2-a^2{x}^2={(a^2-b^2)}^2$
• D. $b^2{y}^2-a^2{x}^2={(a^2+b^2)}^2$

Answer 1

Answer: (B)

$a^2{x}^2-b^2{y}^2={(a^2+b^2)}^2$

Equation of a normal at a parametric point is $a\tan \theta (x-a\sec \theta )+b\sec \theta (y-b\tan \theta )=0$

Thus, $M=(0,\frac{a^2\tan \theta +b^2\tan \theta }{b})$

$N=(\frac{a^2\sec \theta +b^2\sec \theta }{a},0)$

Thus, point $P$ is given by $(\frac{a^2\sec \theta +b^2\sec \theta }{a},\frac{a^2\tan \theta +b^2\tan \theta }{b})$

Now, let $x=\frac{a^2\sec \theta +b^2\sec \theta }{a}$ and $y=\frac{a^2\tan \theta +b^2\tan \theta }{b}$

$\Rightarrow \sec \theta =\frac{ax}{a^2+b^2}$ and $\tan \theta =\frac{by}{a^2+b^2}$

Using the identity ${\sec }^2\theta -{\tan }^2\theta =1$, we get $a^2{x}^2-b^2{y}^2=(a^2+b^2{)}^2$

Hence, option 'B' is correct.