Browse Questions

# 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 $x \sec \theta + y \: cosec \theta -k$, respectively, prove that $p^2+4q^2=k^2$

Toolbox:
• The perpendicular distance (d) of the line $Ax+By+C=0$ from the point $(x_1, y_1)$ is $d = \large\frac{|Ax_1+By_1+C|}{\sqrt {A^2+B^2}}$
The equation of the given lines are
$x \cos \theta - y \sin \theta = k \cos 2 \theta$------------(1)
$x \sec \theta +y \: cosec \theta = k$-------------------(2)
Comparing eqn (1) to the general equation of the line $Ax+By+C=0$ we get,
$A = \cos \theta, B = -\sin \theta \: and \: C = -k \cos 2 \theta$
If $p$ is the perpendicular distance from the origin to line (1) then
$p = \large\frac{|A(0)+B(0)+C|}{\sqrt {A^2+B^2}}$$= \large\frac{|C|}{\sqrt{A^2+B^2}} Substituting the values, p = \large\frac{-k \cos 2 \theta}{\sqrt{\sin^2 \theta+ \cos^2\theta}}$$ = |-k \cos 2 \theta|$-----------(1)
Comparing eqn (2) to the general equation of the line $Ax+By+C=0$ we get,
$A = \sec \theta , B = cosec \theta$, and $C = -k$
$q$ is the length of the perpendicular from the origin (0,0)