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

(N/A) The equation of the given line is $x \cos \alpha + y \sin \alpha = p$.
Rewriting this in the slope-intercept form $y = mx + c$:
$y \sin \alpha = -x \cos \alpha + p$
$y = -x \cot \alpha + \frac{p}{\sin \alpha}$.
Here,the slope $m = -\cot \alpha$ and the intercept $c = \frac{p}{\sin \alpha}$.
We know that the line $y = mx + c$ touches the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ if and only if $c^{2} = a^{2}m^{2} + b^{2}$.
Substituting the values of $m$ and $c$:
$\left(\frac{p}{\sin \alpha}\right)^{2} = a^{2}(-\cot \alpha)^{2} + b^{2}$.
$\frac{p^{2}}{\sin^{2} \alpha} = a^{2} \frac{\cos^{2} \alpha}{\sin^{2} \alpha} + b^{2}$.
Multiplying both sides by $\sin^{2} \alpha$:
$p^{2} = a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha$.
Hence,it is proved that $a^{2} \cos^{2} \alpha + b^{2} \sin^{2} \alpha = p^{2}$.