Question

If $p$ and $q$ are respectively the perpendiculars from the origin upon the straight lines, whose equations are $x\sec \theta +ycosec\theta =a$ and $x\cos \theta -y\sin \theta =a\cos 2\theta$, then $4p^2+q^2$ is equal to

• A. $5a^2$
• B. $4a^2$
• C. $3a^2$
• D. $2a^2$
• E. $a^2$

Answer 1

Answer: (E)

$a^2$

$p^2=\frac{(0\times \sec \theta +0\times cosec\theta -a{)}^2}{{\sec }^2\theta +{\text{cosec}}^2\theta }=\frac{a^2{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }=\frac{a^2{\sin }^{2}2\theta }{4}$

$q^2=\frac{(0\times \cos \theta +0\times \sin \theta -a\cos 2\theta {)}^2}{{\cos }^2\theta +{\sin }^2\theta }=a^2{\cos }^{2}2\theta$

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