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 $\phi$ is given by $ax \sec \phi - by \csc \phi = a^2 -

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$p^2(a^2 \sec^2 \alpha + b^2 \csc^2 \alpha) = (a^2 - b^2)^2$

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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 $\phi$ is given by $ax \sec \phi - by \csc \phi = a^2 - b^2$ ... $(i)$.The given line is $x \cos \alpha + y \sin \alpha = p$.If $(i)$ and the given line represent the same line,then the ratios of their coefficients must be equal:$\frac{a \sec \phi}{\cos \alpha} = \frac{-b \csc \phi}{\sin \alpha} = \frac{a^2 - b^2}{p}$.From this,we get:$\cos \phi = \frac{ap}{(a^2 - b^2) \cos \alpha}$ and $\sin \phi = \frac{-bp}{(a^2 - b^2) \sin \alpha}$.Using the identity $\sin^2 \phi + \cos^2 \phi = 1$,we substitute the values:$\frac{b^2 p^2}{(a^2 - b^2)^2 \sin^2 \alpha} + \frac{a^2 p^2}{(a^2 - b^2)^2 \cos^2 \alpha} = 1$.Multiplying by $(a^2 - b^2)^2$,we get:$p^2 \left( \frac{b^2}{\sin^2 \alpha} + \frac{a^2}{\cos^2 \alpha} \right) = (a^2 - b^2)^2$.Thus,$p^2(a^2 \sec^2 \alpha + b^2 \csc^2 \alpha) = (a^2 - b^2)^2$.

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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