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If the straight-line x \cos\alpha + y \sin\alpha = p touches the curve \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 then prove that a^2 \cos2\alpha + b^2 \sin2\alpha = p^2.

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Given: equation of straight-line x \cos\alpha + y \sin\alpha = p, equation of curve \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 and the straight line touches the curve

To prove: a^2 cos2\alpha + b^2 sin2\alpha = p^2.

Explanation: the know line equation is,

x \cos\alpha + y \sin\alpha = p

\Rightarrow y sin \alpha = p-x cos \alpha

\\\Rightarrow y=\frac{p-x \cos \alpha}{\sin \alpha}$\\\\ $\Rightarrow y=\frac{p}{\sin \alpha}-\frac{x \cos \alpha}{\sin \alpha}$ \\\\$\Rightarrow \mathrm{y}=\frac{\mathrm{p}}{\sin \alpha}-\mathrm{x}\left(\frac{\cos \alpha}{\sin \alpha}\right)$ \\\\$\Rightarrow \mathrm{y}=-\mathrm{x}\left(\frac{\cos \alpha}{\sin \alpha}\right)+\frac{\mathrm{p}}{\sin \alpha}$ \\\\Comparing this with the equation $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ the slope and intercept of the given line can be seen as \\\\$\mathrm{m}=-\frac{\cos \alpha}{\sin \alpha}, \mathrm{C}=\frac{\mathrm{p}}{\sin \alpha}$

We know that, if a line y = mx+c touches the eclipse\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, then required condition is

c\textsuperscript{2} = a\textsuperscript{2}m\textsuperscript{2}+b\textsuperscript{2}

Then putting the corresponding values, we get

\\\left(\frac{p}{\sin \alpha}\right)^{2}=a^{2}\left(-\frac{\cos \alpha}{\sin \alpha}\right)^{2}+b^{2}\\\\$ $\frac{p^{2}}{\sin ^{2} \alpha}=\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}\left(a^{2}\right)+b^{2}\\\\$ $\frac{\mathrm{p}^{2}}{\sin ^{2} \alpha}=\frac{\mathrm{a}^{2} \cos ^{2} \alpha+\mathrm{b}^{2} \sin ^{2} \alpha}{\sin ^{2} \alpha}$
Removing the like terms we get,
$\mathrm{p}^{2}=\mathrm{b}^{2} \sin ^{2} \alpha+\mathrm{a}^{2} \cos ^{2} \alpha$
Hence, proved.

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