If the line x cos alpha + y sin alpha = p is | vedclass

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

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$p^{2}(a^{2} \sec^{2} \alpha + b^{2} \operatorname{cosec}^{2} \alpha) = (a^{2} - b^{2})^{2}$

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(D) The equation of any normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at point $(a \cos \phi, b \sin \phi)$ is given by $ax \sec \phi - by \operatorname{cosec} \phi = a^{2} - b^{2} \quad \dots (i)$Comparing this with the given line $x \cos \alpha + y \sin \alpha = p$,we have:$\frac{a \sec \phi}{\cos \alpha} = \frac{-b \operatorname{cosec} \phi}{\sin \alpha} = \frac{a^{2} - b^{2}}{p}$From this,we get $\sec \phi = \frac{(a^{2} - b^{2}) \cos \alpha}{ap}$ and $\operatorname{cosec} \phi = \frac{-(a^{2} - b^{2}) \sin \alpha}{bp}$.Using the identity $\cos^{2} \phi + \sin^{2} \phi = 1$,we have $\frac{1}{\sec^{2} \phi} + \frac{1}{\operatorname{cosec}^{2} \phi} = 1$.Substituting the values,we get $\frac{a^{2} p^{2}}{(a^{2} - b^{2})^{2} \cos^{2} \alpha} + \frac{b^{2} p^{2}}{(a^{2} - b^{2})^{2} \sin^{2} \alpha} = 1$.Multiplying by $(a^{2} - b^{2})^{2}$,we get $p^{2} (a^{2} \sec^{2} \alpha + b^{2} \operatorname{cosec}^{2} \alpha) = (a^{2} - b^{2})^{2}$.

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$,then

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