Question

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos \theta -y\sin \theta =k\cos 2\theta$ and $x\sec \theta +y\csc \theta =k$, respectively, prove that $p^2+4q^2=k^2$

Answer 1

The equation given lines are $x\cos \theta -y\sin \theta =k\cos 2\theta$....(1)
$x\sec \theta +y\csc \theta =k$....(2)
The perpendicular distance (d) of a line $Ax+By+C=$ $0$ from a point$(x_1,y_1)$is given by $d=\frac{|A{x}_1+B{y}_1+C|}{\sqrt{A^2+B^2}}$
On comparing equation(1) to the general equation of line i.e; $Ax+By+C$ $=$0, we obtain $A=\cos \theta ,B=-\sin \theta ,$ and $C=-k\cos 2\theta$
It is given that p is the length of the perpendicular from(0,0) to line (1)
$\therefore p=\frac{\left|A\right(0)+B(0)+C|}{\sqrt{A^2+B^2}}=\frac{\left|C\right|}{\sqrt{A^2+B^2}}$ $=\frac{|-k\cos 2\theta |}{\sqrt{{\cos }^2\theta +{\sin }^2\theta }}=|-k\cos 2\theta |$......(3)
On comparing equation(2) to the general equation of line i.e; $Ax+By+C=0$, we obtain $A=\sec \theta ,B=\csc \theta ,$ and $C=-k$
It is given that $q$ is the length of the perpendicular from $(0,0)$ to line (2)
$\therefore q=\frac{\left|A\right(0)+B(0)+C|}{\sqrt{A^2+B^2}}=\frac{\left|C\right|}{\sqrt{A^2+B^2}}=\frac{|-k|}{\sqrt{{\sec }^2\theta +{\csc }^2\theta }}$.....(4)
From (3) and (4) we have
$p^2+4q^2={\left(|-k\cos 2\theta |\right)}^2+4\left\{\frac{|-k|}{\sqrt{{\sec }^2\theta +{\csc }^2\theta }}\right\}$
$=k^2{\cos }^{2}2\theta +\frac{4k^2}{({\sec }^2\theta +{\csc }^2\theta )}$
$=k^2{\cos }^{2}2\theta +\frac{4k^2}{\{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }\}}$
$=k^2{\cos }^{2}2\theta +\frac{4k^2}{\left\{\frac{{\sin }^2+{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }\right\}}$
$=k^2{\cos }^{2}2\theta +4k^2{\sin }^2\theta {\cos }^2\theta$
$=k^2{\cos }^{2}2\theta +k^2{\left(2\sin \theta \cos \theta \right)}^2$
$=k^2{\cos }^{2}2\theta +k^2{\sin }^{2}2\theta$
$=k^2({\cos }^{2}2\theta +{\sin }^{2}2\theta )=k^2$
Hence, we proved that $p^2+4q^2=k^2$