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  \(p ^2 + 4q ^2 = k^ 2\).