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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
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The correct answer is:
$\left(a^2+b^2\right)^2$
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 $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2$.
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
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 $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\left(a^2+b^2\right)^2$.
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