#### 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 $a^2 \cos2\alpha + b^2 \sin2\alpha = p^2.$

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,

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

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

\begin{align*} &\Rightarrow y = \frac{p - x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - \frac{x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - x\left(\frac{\cos \alpha}{\sin \alpha}\right) \\ &\Rightarrow y = -x\left(\frac{\cos \alpha}{\sin \alpha}\right) + \frac{p}{\sin \alpha} \\ &\text{Comparing this with the equation } y = mx + c \text{, the slope and intercept of the given line can be seen as} \\ &m = -\frac{\cos \alpha}{\sin \alpha}, \quad c = \frac{p}{\sin \alpha} \end{align*}

\begin{align*} &\Rightarrow y = \frac{p - x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - \frac{x \cos \alpha}{\sin \alpha} \\ &\Rightarrow y = \frac{p}{\sin \alpha} - x\left(\frac{\cos \alpha}{\sin \alpha}\right) \\ &\Rightarrow y = -x\left(\frac{\cos \alpha}{\sin \alpha}\right) + \frac{p}{\sin \alpha} \\ &\text{Comparing this with the equation } y = mx + c \text{, the slope and intercept of the given line can be seen as} \\ &m = -\frac{\cos \alpha}{\sin \alpha}, \quad c = \frac{p}{\sin \alpha} \end{align*}

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

$\left(\frac{p}{\sin \alpha}\right)^{2} = a^{2}\left(-\frac{\cos \alpha}{\sin \alpha}\right)^{2} + b^{2} \\ \frac{p^{2}}{\sin^{2} \alpha} = \frac{\cos^{2} \alpha}{\sin^{2} \alpha}\left(a^{2}\right) + b^{2} \\ \frac{p^{2}}{\sin^{2} \alpha} = \frac{a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha}{\sin^{2} \alpha}$

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