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A tangent to the hyperbola $\frac{x^2}{4}-\frac{y^2}{2}=1$ meets $x$-axis at $\mathrm{P}$ and $y$-axis at $\mathrm{Q}$. Lines $\mathrm{PR}$ and $\mathrm{QR}$ are drawn such that OPRQ is a rectangle (where $\mathrm{O}$ is the origin). Then $\mathrm{R}$ lies on :
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Verified Answer
The correct answer is:
$\frac{4}{x^2}-\frac{2}{y^2}=1$
$\frac{4}{x^2}-\frac{2}{y^2}=1$
Equation of the tangent at the point ' $\theta$ ' is
$$
\begin{aligned}
& \frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1 \\
& \Rightarrow P=(a \cos \theta, 0) \text { and } Q=(0,-b \cot \theta) \\
& \text { Let } \mathrm{R} \text { be }(\mathrm{h}, \mathrm{k}) \Rightarrow \mathrm{h}=a \cos \theta, \mathrm{k}=-b \cot \theta \\
& \Rightarrow \quad \frac{k}{h}=\frac{-b}{a \sin \theta} \Rightarrow \sin \theta=\frac{-b h}{a k} \quad \text { and } \\
& \cos \theta=\frac{h}{a}
\end{aligned}
$$
By squaring and adding,
$$
\frac{b^2 h^2}{a^2 k^2}+\frac{h^2}{a^2}=1
$$
$$
\begin{aligned}
& \Rightarrow \frac{b^2}{k^2}+1=\frac{a^2}{h^2} \\
& \Rightarrow \frac{a^2}{h^2}-\frac{b^2}{k^2}=1
\end{aligned}
$$
Now, given $\mathrm{q}^{\mathrm{n}}$ 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., $\frac{4}{x^2}-\frac{2}{y^2}=1$
$$
\begin{aligned}
& \frac{x \sec \theta}{a}-\frac{y \tan \theta}{b}=1 \\
& \Rightarrow P=(a \cos \theta, 0) \text { and } Q=(0,-b \cot \theta) \\
& \text { Let } \mathrm{R} \text { be }(\mathrm{h}, \mathrm{k}) \Rightarrow \mathrm{h}=a \cos \theta, \mathrm{k}=-b \cot \theta \\
& \Rightarrow \quad \frac{k}{h}=\frac{-b}{a \sin \theta} \Rightarrow \sin \theta=\frac{-b h}{a k} \quad \text { and } \\
& \cos \theta=\frac{h}{a}
\end{aligned}
$$
By squaring and adding,
$$
\frac{b^2 h^2}{a^2 k^2}+\frac{h^2}{a^2}=1
$$
$$
\begin{aligned}
& \Rightarrow \frac{b^2}{k^2}+1=\frac{a^2}{h^2} \\
& \Rightarrow \frac{a^2}{h^2}-\frac{b^2}{k^2}=1
\end{aligned}
$$
Now, given $\mathrm{q}^{\mathrm{n}}$ 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., $\frac{4}{x^2}-\frac{2}{y^2}=1$
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