Question

A tangent to the circle $x^2+y^2=4$ intersects the hyperbola $x^2-2y^2=2$ at $P$ and $Q$. If locus of mid-point of $PQ$ is ${\left(x^2-2y^2\right)}^2=\lambda \left(x^2+4y^2\right)$, 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
$\frac{hx}{2}-ky=\frac{h^2}{2}-\frac{k^2}{1}\text{ (using }T=S_1\text{ ) }$
As, it is tangent to the circle $x^2+y^2=4$, so
$\left|\frac{\frac{h^2}{2}-k^2}{\sqrt{\frac{h^2}{4}+k^2}}\right|=2$
$\Rightarrow {\left(\frac{h^2}{2}-k^2\right)}^2=4\left(\frac{h^2}{4}+k^2\right)$
$\Rightarrow \text{ Locus of }(h,k)\text{ is }{\left(x^2-2y^2\right)}^2=4\left(x^2+4y^2\right)$
$\therefore \lambda =4$