The line x cos alpha + y sin alpha = p will b | vedclass

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(C) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be rewritten as $y = -x \cot \alpha + \frac{p}{\sin \alpha}$.Comparing t

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$p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$

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(C) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be rewritten as $y = -x \cot \alpha + \frac{p}{\sin \alpha}$.Comparing this with the line $y = mx + c$,we have $m = -\cot \alpha$ and $c = \frac{p}{\sin \alpha}$.The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $c^2 = a^2 m^2 + b^2$.Substituting the values of $m$ and $c$:$(\frac{p}{\sin \alpha})^2 = a^2(-\cot \alpha)^2 + b^2$$\frac{p^2}{\sin^2 \alpha} = a^2 \cot^2 \alpha + b^2$$p^2 = a^2 \cot^2 \alpha \sin^2 \alpha + b^2 \sin^2 \alpha$$p^2 = a^2 \cos^2 \alpha + b^2 \sin^2 \alpha$.

The line $x \cos \alpha + y \sin \alpha = p$ will be a tangent to the conic $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,if

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