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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
$\frac{4}{{{x^2}}} + \frac{2}{{{y^2}}} = 1$
$\frac{2}{{{x^2}}} - \frac{4}{{{y^2}}} = 1$
$\frac{2}{{{x^2}}} + \frac{4}{{{y^2}}} = 1$
$\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1$
Solution
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)\,\,\,\,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}}$
$\cos \theta = \frac{h}{a}$
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 $e{q^n}$ 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., $\frac{4}{{{x^2}}} – \frac{2}{{{y^2}}} = 1$
