Let $\lambda x - 2y = \mu$ be a tangent to the hyperbola ${a^2}{x^2} - {y^2} = {b^2}$. Then ${\left( {{\lambda \over a}} \right)^2} - {\left( {{\mu \over b}} \right)^2}$ is equal to :
Solution
$\frac{x^{2}}{\left(\frac{b^{2}}{a^{2}}\right)}-\frac{y^{2}}{b^{2}}=1$
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Tangent in slope form $\Rightarrow y=m x \pm \sqrt{\frac{b^{2}}{a^{2}} m^{2}-b^{2}}$
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i.e., same as $y=\frac{\lambda x}{2}-\frac{\mu}{2}$
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Comparing coefficients,
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$$
\begin{aligned}
&m=\frac{\lambda}{2}, \frac{b^{2}}{a^{2}} m^{2}-b^{2}=\frac{\mu^{2}}{4} \\\\
&\text { Eliminating } m, \frac{b^{2}}{a^{2}} \cdot \frac{\lambda^{2}}{4}-b^{2}=\frac{\mu^{2}}{4} \\\\
&\Rightarrow \frac{\lambda^{2}}{a^{2}}-\frac{\mu^{2}}{b^{2}}=4
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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