Let [ x - 2y = ] be a tangent to the hyperbola a^2x^2 - y^2 = b^2.

Question

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 :

Accepted Answer

$\frac{x^{2}}{\left(\frac{b^{2}}{a^{2}}\right)}-\frac{y^{2}}{b^{2}}=1$

Tangent in slope form $\Rightarrow y=m x \pm \sqrt{\frac{b^{2}}{a^{2}} m^{2}-b^{2}}$

i.e., same as $y=\frac{\lambda x}{2}-\frac{\mu}{2}$

Comparing coefficients,

$$
\begin{aligned}
&m=\frac{\lambda}{2}, \frac{b^{2}}{a^{2}} m^{2}-b^{2}=\frac{\mu^{2}}{4} \\
&\text { Eliminating } m, \frac{b^{2}}{a^{2}} \cdot \frac{\lambda^{2}}{4}-b^{2}=\frac{\mu^{2}}{4} \\
&\Rightarrow \frac{\lambda^{2}}{a^{2}}-\frac{\mu^{2}}{b^{2}}=4
\end{aligned}
$$

Solutions · 1