Question

A tangent to the circle $x^2+y^2=4$ intersects the hyperbola $x^2-2 y^2=2$ at $P$ and $Q$. If locus of mid-point of $P Q$ is $\left(x^2-2 y^2\right)^2=\lambda\left(x^2+4 y^2\right)$, then $\lambda$ equals

• A. 4
• B. 2
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

Equation of chord of hyperbola $\frac{x^2}{2}-\frac{y^2}{1}=1$, whose mid-point is $(h, k)$ is
$\frac{h x}{2}-k y=\frac{h^2}{2}-\frac{k^2}{1} \text { (using } T=S_1 \text { ) }$
As, it is tangent to the circle $x^2+y^2=4$, so
$\left|\frac{\frac{ h ^2}{2}- k ^2}{\sqrt{\frac{ h ^2}{4}+ k ^2}}\right|=2$
$\Rightarrow\left(\frac{ h ^2}{2}- k ^2\right)^2=4\left(\frac{ h ^2}{4}+ k ^2\right)$
$\Rightarrow \text { Locus of }( h , k ) \text { is }\left( x ^2-2 y ^2\right)^2=4\left( x ^2+4 y ^2\right)$
$\therefore \lambda=4 $