Show that the product of the perpendiculars drawn from the two points $(\pm \sqrt {a^2-b^2} , 0)$ upon the line $\dfrac {x}{a} \cos \theta + \dfrac {y}{b} \sin \theta = 1$ is $b^2$.
My Attempt: Let $p_1$ and $p_2$ be the lengths of perpendiculars. Then, $$p_1=|\dfrac {b\cos \theta (-\sqrt {a^2-b^2}) - ab}{\sqrt {a^2\sin^2 \theta + b^2 \cos^2 \theta}}|$$ And, $$p_2=|\dfrac {(b\cos \theta)(\sqrt {a^2-b^2}) - ab}{\sqrt {a^2\sin^2 \theta + b^2\cos^2 \theta}}|$$ On Multiplying these two I got, $$=\dfrac {-a^2b^2 \cos^2 \theta+b^4\cos^2 \theta - a^2b^2}{a^2\sin^2 \theta + b^2\cos^2 \theta}$$
Check the sign of the $a^2b^2$ term in the numerator: it should be opposite to the sign of $a^2b^2\cos^2{\theta}$. Then use $\cos^2{\theta}+\sin^2{\theta}=1$ to cancel.