If the line lx + my = 1 is a normal to the hy | vedclass

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(C) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $\fr

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$(a^2 + b^2)^2$

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(C) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2 + b^2$. Comparing this with the given line $lx + my = 1$,we can rewrite the normal equation as $\frac{a \cos 	heta}{a^2 + b^2} x + \frac{b \cot 	heta}{a^2 + b^2} y = 1$. Thus,$l = \frac{a \cos 	heta}{a^2 + b^2}$ and $m = \frac{b \cot 	heta}{a^2 + b^2}$. Now,$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{a^2 (a^2 + b^2)^2}{a^2 \cos^2 	heta} - \frac{b^2 (a^2 + b^2)^2}{b^2 \cot^2 	heta}$. $= (a^2 + b^2)^2 (\sec^2 	heta - 	an^2 	heta)$. Since $\sec^2 	heta - 	an^2 	heta = 1$,we get $\frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2$.

If the line $lx + my = 1$ 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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