If $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $a^2 m^2-b^2 l^2=$
If $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $a^2 m^2-b^2 l^2=$
$\frac{m^2}{l^2}\left(a^2+b^2\right)^2$
$\left(l^2+m^2\right)\left(a^2+b^2\right)^2$
$\frac{l^2}{m^2}\left(a^2+b^2\right)^2$
$l^2 m^2\left(a^2+b^2\right)^2$
Solution
Given equation of normal, $l x+m y=1$
$
y=\frac{-l}{m}+\frac{1}{m}
$
Which is of the form $y=m x+c$
Since, the lines $y=m x+c$ will be normal to the hyperbola
$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \text { if } c^2=\frac{m^2\left(a^2+b^2\right)^2}{a^2-b^2 m^2} \\
& \therefore \quad\left(\frac{1}{m}\right)^2=\frac{\left(-\frac{l}{m}\right)^2\left(a^2+b^2\right)^2}{a^2-b^2\left(-\frac{l}{m}\right)^2}
\end{aligned}
$
$\begin{aligned} & \frac{1}{m^2}=\frac{\frac{l^2}{m^2}\left(a^2+b^2\right)^2}{\frac{m^2 a^2-l^2 b^2}{m^2}} \\ & \frac{1}{m^2}=\frac{l^2\left(a^2+b^2\right)^2}{m^2 a^2-l^2 b^2} \\ & a^2 m^2-b^2 l^2=l^2 m^2\left(a^2+b^2\right)^2\end{aligned}$