Question

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

Answer 1

The equations of given lines are $x \space cos θ \- y \space sinθ = k \space cos 2θ … (1)$

$x\space secθ + y \space cosec θ= k … (2)$
The perpendicular distance (d) of a line $Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}$
On comparing equation (1) to the general equation of line i.e., $Ax + By + C = 0$, we obtain $A = cosθ, B = -sinθ,$ and $C = -k\space cos 2θ.$

It is given that p is the length of the perpendicular from $(0, 0)$ to line (1).

$∴ p=\frac{\left|A(0)+B(0)+C\right|}{\sqrt{A^2+B^2}}$

$=\frac{\left|C\right|}{\sqrt{A^2+B^2}}$

$=\frac{\left|-k cos 2θ\right|}{\sqrt{cos^2θ+sin^2θ}}$

$=\left|-kcos2θ\right|....(3)$

On comparing equation (2) to the general equation of line i.e., $Ax + By + C = 0$, we obtain $A = secθ, B = cosecθ,$ and $C = -k.$
It is given that q is the length of the perpendicular from $(0, 0)$ to line (2).

$∴ q=\frac{\left|A(0)+B(0)+C\right|}{\sqrt{A^2+B^2}}$

$=\frac{\left|C\right|}{\sqrt{A^2+B^2}}$

$=\frac{\left|-k\right|}{\sqrt {sec^2θ+cosec^2θ}}........(4)$

From (3) and (4), we have
$p^2 \+ 4q^2 =\left(\left|-kcos2θ\right|\right)^2+4\left(\frac{\left|-k\right|}{\sqrt{sec^2θ+cosec^2θ}}\right)^2$

$=k^2cos^2 2θ+\frac{4k^2}{\left(sec^2θ+cosec^2θ\right)}$

$=k^2cos^2 2θ+\frac{4k^2}{\left(\frac{1}{cos^2θ}+\frac{1}{sin^2θ}\right)}$

$=k^2cos^2 2θ+\frac{4k^2}{\left(\frac{sin2θ+cos2θ}{sin^2θcos^2θ}\right)}$

$=k^2cos^2 2θ+\frac{4k^2}{\left(\frac{1}{sin^2θcos^2θ}\right)}$

$=k^2cos^2 2θ+4k^2sin^2θcos^2θ$

$=k^2cos^2 2θ+k^2sin^2 2θ$

$=k^2(cos^2 2θ+sin^2 2θ)$

$=k^2$

Hence, we proved that $p ^2 + 4q ^2 = k^ 2$.