Question

If p and q are length of the perpendiculars from the origin on the lines $x\sec \theta +y\csc \theta =aandx\cos \theta -y\sin \theta =a\cos 2\theta ,then4p^2+q^2$ equals

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

Answer 1

The correct option is B $a^2$

Given lines

$x\sec \theta +y\csc \theta =a----\left(1\right)$

$x\cos \theta -y\sin \theta =a\cos 2\theta ----\left(2\right)$

On comparing both equation with $ax+by+c=0$ we get

$a_1=\sec \theta ,b_1=\csc \theta ,c=-a$

$a_2=\cos \theta ,b_2=-\sin \theta ,c=-a\cos 2\theta$

p is the length of perpendicular from origin to line $x\sec \theta +y\csc \theta =a$

By perpendicular distance formula $p=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

$p=\left|\frac{0+0-a}{\sqrt{{\sec }^2\theta +{\csc }^2\theta }}\right|$

$p=\left|\frac{-a}{\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}}\right|$

$p=\frac{a}{\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }}}$

But ${\sin }^2\theta +{\cos }^2\theta =1$

$p=\frac{a\sin 2\theta }{2}$

q is the length of perpendicular from origin to line $x\cos \theta -y\sin \theta =a\cos 2\theta$

By perpendicular distance formula $q=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

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

$q=\frac{a\cos 2\theta }{1}$

$q=a\cos 2\theta$

$4p^2+q^2=4{\left(\frac{a\sin 2\theta }{2}\right)}^2+(a\cos 2\theta {)}^2$

$4p^2+q^2=4\left(\frac{a^2{\sin }^{2}2\theta }{4}\right)+a^2{\cos }^{2}2\theta$

$4p^2+q^2=a^2{\sin }^{2}2\theta +a^2{\cos }^{2}2\theta$

$4p^2+q^2=a^2({\sin }^{2}2\theta +{\cos }^{2}2\theta )$

$4p^2+q^2=a^2$