The line lx + my + n = 0 is a normal to the e | vedclass

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(B) The equation of any normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is given by $ax \sec \

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$\frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2 - b^2)^2}{n^2}$

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(B) The equation of any normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(a \cos 	heta, b \sin 	heta)$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$ $(i)$.The given line is $lx + my + n = 0$ $(ii)$.If $(i)$ and $(ii)$ represent the same line,then the ratios of their coefficients must be equal:$\frac{a \sec 	heta}{l} = \frac{-b \csc 	heta}{m} = \frac{-(a^2 - b^2)}{n}$.From this,we get:$\cos 	heta = \frac{-an}{l(a^2 - b^2)}$ and $\sin 	heta = \frac{bn}{m(a^2 - b^2)}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we have:$\frac{a^2 n^2}{l^2(a^2 - b^2)^2} + \frac{b^2 n^2}{m^2(a^2 - b^2)^2} = 1$.Multiplying both sides by $(a^2 - b^2)^2 / n^2$,we get:$\frac{a^2}{l^2} + \frac{b^2}{m^2} = \frac{(a^2 - b^2)^2}{n^2}$.

The line $lx + my + n = 0$ is a normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

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