Question

A tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ meets $x$ -axis at $P$ and $y$ -axis at $Q$ . Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on :

• A. $\frac{4}{x^2}+\frac{2}{y^2}=1$
• B. $\frac{2}{x^2}-\frac{4}{y^2}=1$
• C. $\frac{2}{x^2}+\frac{4}{y^2}=1$
• D. $\frac{4}{x^2}-\frac{2}{y^2}=1$

Answer 1

Answer: (D)

Equation of the tangent at the point $\theta$ is $\frac{xsec\theta }{a}-\frac{ytan\theta }{b}=1$
$\Rightarrow P=\left(acos\theta ,0\right)$ and $Q=\left(0,-bcot\theta \right)$
Let $R$ be $\left(h,k\right)\Rightarrow h=acos\theta ,k=-bcot\theta$
$\Rightarrow \frac{k}{h}=\frac{-b}{asin\theta }\Rightarrow sin\theta =\frac{-bh}{ak}$ and $cos\theta =\frac{h}{a}$
By squaring and adding,
$\frac{b^2{h}^2}{a^2{k}^2}+\frac{h^2}{a^2}=1$

$\Rightarrow \frac{b^2}{k^2}+1=\frac{a^2}{h^2}\Rightarrow \frac{a^2}{h^2}-\frac{b^2}{k^2}=1$
Now, given equation of hyperbola is
$\frac{x^2}{4}-\frac{y^2}{2}=1\Rightarrow a^2=4,b^2=2$
$\therefore R$ lies on $\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$ i.e., $\frac{4}{x^2}-\frac{2}{y^2}=1$