If the line $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then…
If the line $l x+m y=1$ is a normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $\frac{a^2}{l^2}-\frac{b^2}{m^2}$ is equal to
$a^2-b^2$
$a^2+b^2$
$\left(a^2+b^2\right)^2$
$\left(a^2-b^2\right)^2$
Solution
If $l x+m y+n=0$ is normal to the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Then $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{\left(a^2+b^2\right)^2}{n^2}$
Here, $n=-1$, therefore $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2$.