Question

A tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ cuts the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in points $P$ & $Q$. The locus of the mid- point $PQ$ is

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

Answer 1

The correct option is A $(\frac{x^2}{a^2}-\frac{y^2}{b^2})={(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2$

Any tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
is $\frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1\cdots \left(1\right)$
Let the mid - point of $PQ$ be $(h,k)$
Then the equation of the chord bisected at $(h,k)$ to the ellipse is
$T=S_1\Rightarrow \frac{hx}{a^2}+\frac{ky}{b^2}=\frac{h^2}{a^2}+\frac{k^2}{b^2}\cdots \left(2\right)$

$\because$ Equation $\left(1\right)$ & $\left(2\right)$ are identical.
$\therefore \frac{x_1}{h}=-\frac{y_1}{k}=\frac{1}{\frac{h^2}{a^2}+\frac{k^2}{b^2}}$
$x_1=\frac{h}{(\frac{h^2}{a^2}+\frac{k^2}{b^2})},y_1=\frac{-k}{(\frac{h^2}{a^2}+\frac{k^2}{b^2})}$

Putting the values of $(x_1,y_1)$ in the hyperbola, we get
$\frac{h^2}{a^2}-\frac{k^2}{b^2}={(\frac{h^2}{a^2}+\frac{k^2}{b^2})}^2$
Hence required locus will be
$\frac{x^2}{a^2}-\frac{y^2}{b^2}={(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2$