If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin to the straight lines $x \sec \theta + y \text{cosec } \theta = 2a$ and $x \cos \theta - y \sin \theta = a \cos 2 \theta$, then prove that $p^2_1+p^2_2 = a^2$.
The given equations can be written as
$x \sec \theta + y \text{cosec } \theta - 2a = 0$ $\;\;\; \cdots \; (1)$
and $x \cos \theta - y \sin \theta - a \cos 2 \theta = 0$ $\;\;\; \cdots \; (2)$
Length of perpendicular from $\left(0,0\right)$ to equation $(1)$ is
$p_1 = \left|\dfrac{-2a}{\sqrt{\sec^2 \theta + \text{cosec}^2 \theta}}\right|$
i.e. $p^2_1 = \dfrac{4a^2}{\dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta}}$
i.e. $p^2_1 = \dfrac{4a^2 \times \sin^2 \theta \cos^2 \theta}{\sin^2 \theta + \cos^2 \theta}$
i.e. $p^2_1 = 4 a^2 \sin^2 \theta \cos^2 \theta$ $\;\;\; \cdots \; (3)$
Length of perpendicular from $\left(0,0\right)$ to equation $(2)$ is
$p_2 = \left|\dfrac{- a \cos 2 \theta}{\sqrt{\left(\cos \theta\right)^2 + \left(- \sin \theta\right)^2}}\right|$
i.e. $p^2_2 = a^2 \cos^2 2 \theta$
i.e. $p^2_2 = a^2 \left(\cos^2 \theta - \sin^2 \theta\right)^2$
i.e. $p^2_2 = a^2 \left(\cos^4 \theta + \sin^4 \theta - 2 \cos^2 \theta \sin^2 \theta\right)$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ From equations $(3)$ and $(4)$,
$\begin{aligned}
p^2_1 + p^2_2 & = 4 a^2 \sin^2 \theta \cos^2 \theta + a^2 \cos^4 \theta + a^2 \sin^4 \theta - 2a^2 \cos^2 \theta \sin^2 \theta \\\\
& = a^2 \left(\cos^4 \theta + 2a^2 \cos^2 \theta \sin^2 \theta + \sin^4 \theta\right) \\\\
& = a^2 \left(\cos^2 \theta + \sin^2 \theta\right)^2 \\\\
& = a^2
\end{aligned}$
i.e. $p^2_1 + p^2_2 = a^2$ $\;\;\;$ Hence proved