Question

Q. A tangent to the hyperbola \(\frac{x^{4}}{4}-\frac{y^{2}}{2}=1\) meets \(\mathrm{x}\)-axis at \(P\) and \(\mathrm{y}\)-axis at \(Q .\) Lines \(P R\) and \(Q R\) are drawn such that \(O P R Q\) 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

Solution:
Equation of the tangent at the point ' \(\theta\) ' is
$$
\frac{x \operatorname{sex} \theta}{a}-\frac{x \tan \theta}{b}=1
$$
$$
\Rightarrow P=(a \cos \theta, 0) \text { and } Q=(0,-b \cot \theta)
$$
Let \(R\) be \((h, k) \Rightarrow h=a \cos \theta, k=-b \cot \theta\)
$$
\Rightarrow \frac{k}{h}=\frac{-b}{a \sin \theta} \Rightarrow \sin \theta=\frac{b h}{a k} \text { and }
$$
$$
\cos \theta=\frac{h}{a}
$$
By squaring and adding,
$$
\begin{aligned}
&\frac{b^{2} h^{2}}{a^{2} k^{2}}+\frac{h^{2}}{a^{2}}=1 \\
&\Rightarrow \frac{b^{2}}{k^{2}}+1=\frac{a^{2}}{h^{2}} \\
&\Rightarrow \frac{a^{2}}{h^{2}}-\frac{b^{2}}{k^{2}}=1
\end{aligned}
$$
Now, given \(e q^{n}\) of hyperbola is \(\frac{x^{2}}{4}-\frac{y^{2}}{2}=1\)
$$
\Rightarrow a^{2}=4, b^{2}=2
$$
$$
\therefore R \text { lies on } \frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}=1 i . e ., \frac{4}{x^{2}}-\frac{2}{y^{2}}=1
$$

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