Question

If $p_1$, $p_2$ denote the lengths of the perpendiculars from the origin on the lines $x\sec \alpha +y\text{cosec}\alpha =2a$ and $x\cos \alpha +y\sin \alpha =a\cos 2\alpha$ respectively, then ${(\frac{p_1}{p_2}+\frac{p_2}{p_1})}^2$ is equal to

• A. $4{\sin }^{2}4\alpha$
• B. $4{\cos }^{2}4\alpha$
• C. $4{\text{cosec}}^{2}4\alpha$
• D. $4{\sec }^{2}4\alpha$

Answer 1

The correct option is C $4{\text{cosec}}^{2}4\alpha$

Given lines

$x\sec \alpha +ycosec\alpha =2a$-----(1)

$x\cos \alpha +y\sin \alpha =a\cos 2\alpha$------(2)

$p_1$ is the perpendicular lengths from origin from (1)

$p_1=\left|\frac{-2a}{\sqrt{{\sec }^2\alpha +cose{c}^2\alpha }}\right|$

$p_1=\frac{2a}{\sqrt{\frac{1}{{\cos }^2\alpha }+\frac{1}{{\sin }^2\alpha }}}$

$p_1=\frac{2a}{\sqrt{\frac{{\sin }^2\alpha +{\cos }^2\alpha }{{\sin }^2\alpha {\cos }^2\alpha }}}$

$p_1=\frac{2a}{\sqrt{\frac{1}{{\sin }^2\alpha {\cos }^2\alpha }}}$

$p_1=\frac{2a}{\frac{1}{\sin \alpha \cos \alpha }}$

$p_1=a2\sin \alpha \cos \alpha$

$p_1=a\sin 2\alpha$

$p_2$ is the perpendicular lengths from origin from (2)

$p_2=\left|\frac{-a\cos 2\alpha }{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}\right|$

$p_2=\frac{a\cos 2\alpha }{\sqrt{1}}$

$p_2=a\cos 2\alpha$

$\frac{p_1}{p_2}=\frac{a\sin 2\alpha }{a\cos 2\alpha }$

$\frac{p_1}{p_2}=\tan 2\alpha$---(3)

$\frac{p_2}{p_1}=\frac{a\cos 2\alpha }{a\sin 2\alpha }$

$\frac{p_2}{p_1}=\cot 2\alpha$----(4)

Adding eq (3) and (4)

$\tan 2\alpha +\cot 2\alpha$

$\frac{\sin 2\alpha }{\cos 2\alpha }+\frac{\cos 2\alpha }{\sin 2\alpha }$

$\frac{{\sin }^{2}2\alpha +{\cos }^{2}2\alpha }{\cos 2\alpha \sin 2\alpha }$

$\frac{1}{\cos 2\alpha \sin 2\alpha }$

on mutiplying and dividing by 2

$\frac{2}{2\cos 2\alpha \sin 2\alpha }$

$\frac{2}{\sin 4\alpha }$

${(\frac{p_1}{p_2}+\frac{p_2}{p_1})}^2={\left(\frac{2}{\sin 4\alpha }\right)}^2$

${(\frac{p_1}{p_2}+\frac{p_2}{p_1})}^2=\frac{4}{{\sin }^{2}4\alpha }=4cose{c}^{2}4\alpha$