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 $\frac{x sec \theta }{a}-\frac{y tan \theta }{b}=1$
$\Rightarrow P=\left(a cos \theta , 0\right)$ and $Q=\left(0 , - b cot \theta \right)$
Let $R$ be $\left(h , k\right)\Rightarrow h=acos\theta ,k=-bcot\theta $
$\Rightarrow \frac{k}{h}=\frac{- b}{a sin \theta }\Rightarrow sin\theta =\frac{- b h}{a k}$ 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., $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$