Question

If $p$ and $p'$ be the perpendiculars from the origin upon the straight lines $x\, sec\, \theta + y \,cosec \,\theta = a$ and $x \,cos\, \theta - y \,sin \,\theta = a\, cos \,2\theta$ respectively, then the value of the expression $4p^2 + p'^{2}$ is

• A. $a^2$
• B. $3a^2$
• C. $2a^2$
• D. $4a^2$

Answer 1

Answer: (A)

We have $p =$ length of the perpendicular from $(0,0)$ on $x\,sec\, \theta +y \,cosec\, \theta = a$
i.e., $p=\frac{a}{\sqrt{sec^{2}\,\theta+cosec^{2}\,\theta}}=\frac{a\,sin\,2\theta}{2}$
$\Rightarrow 2p = a \,sin \,2\theta\quad\ldots \left(i\right)$
$p' =$ length of the perpendicular from $\left(0, 0\right)$ on
$x\,cos\,\theta-y\,sin\,\theta=a\,cos\,2\theta$
i.e., $p'=\frac{a\,cos\,2\theta}{\sqrt{cos^{2}\,\theta+sin^{2}\,\theta}}=a\,cos\,2\theta\quad\ldots\left(ii\right)$
Squaring and adding $\left(i\right)$ and $\left(ii\right)$, we get
$4p^{2}+p'^{2}=a^{2}$