If $P_{1}$ and $P_{2}$ be the length of perpendiculars from the origin upon the straight lines $x \sec…
If $P_{1}$ and $P_{2}$ be the length of perpendiculars from the origin upon the straight lines $x \sec \theta+y \operatorname{cosec} \theta$ $=a$ and $x \cos \theta-y \sin \theta=a \cos 2 \theta$ respectively,
then the value of $4 \mathrm{P}_{1}^{2}+\mathrm{P}_{2}^{2}$
$a^{2}$
$2 a^{2}$
$a^{2} / 2$
$3 a^{2}$
Solution
We have $P_{1}=$ length of perpendicular from (0,0) on $x \sec \theta+y \operatorname{cosec} \theta=a$
$\begin{aligned} \text { i.e. } P_{1} &=\left|\frac{a}{\sqrt{\sec ^{2} \theta+\cos e c^{2} \theta}}\right|=|a \sin \theta \cos \theta| \\ &=\left|\frac{a}{2} \sin 2 \theta\right| \text { or } 2 P_{1}=|a \sin 2 \theta| \end{aligned}$
$\mathrm{P}_{2}=$ Length of the perpendicular from (0,0) on $x \cos \theta-y \sin \theta=a \cos 2 \theta$
$P_{2}=\left|\frac{a \cos 2 \theta}{\sqrt{\cos ^{2} \theta+\sin ^{2} \theta}}\right|=|a \cos 2 \theta|$
Now, $4 P_{1}^{2}+P_{2}^{2}=a^{2} \sin ^{2} 2 \theta+a^{2} \cos ^{2} 2 \theta=a^{2}$