Question

A tangent to the hyperbola $\frac{x^{4}}{4}-\frac{y^{2}}{2}=1$ 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

The Correct Option is D

Equation of the tangent at the point $?\theta$? is
$\frac{x\,sex\,\theta}{a}-\frac{x\,tan\,\theta }{b}=1$
$\Rightarrow P = \left(a\,cos\,\theta, 0\right)$ and $Q = \left(0, -b\,cot \,\theta\right)$
Let $R$ be $\left(h, k\right) \Rightarrow h = a\,cos\, \theta, k = -b\, cot\,\theta$
$\Rightarrow \frac{k}{h}=\frac{-b}{a\,sin\,\theta} \Rightarrow sin\,\theta=\frac{-bh}{ak}$ 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 $eq^{n}$ of hyperbola is $\frac{x^{2}}{4}-\frac{y^{2}}{2}=1$
$\Rightarrow a ^{2} = 4, b^{2}=2$
$\therefore R$ lies on $\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}=1 i.e., \frac{4}{x^{2}}-\frac{2}{y^{2}}=1$