Question

Prove that the product of the lengths of the perpendiculars drawn from the points \(\left(\sqrt{a^2-b^2},0\right)\) and \(\left(-\sqrt{a^2-b^2},0\right)\) to the line \(\frac{x}{a}cosθ+\frac{y}{b}sinθ=1\) is \(b^2\).

Answer 1

The equation of the given line is \(\frac{x}{a} cosθ+\frac{y}{b}sinθ=1\)

or \(bx\space cosθ+ay\space sinθ-ab=0 .....(1)\)

Length of the perpendicular from point \(\left(\sqrt{a^2-b^2},0\right)\) to line (1) is 

\(P_1=\frac{\left|bcosθ(\sqrt{a^2-b^2},0)+asinθ(0)-ab\right|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}\)

\(=\frac{\left|bcosθ\sqrt{a^2-b^2}-ab\right|}{\sqrt{b^2cos^2θ+a^2sin^2θ }}.....(2)\)

Length of the perpendicular from point \(\left(-\sqrt{a^2-b^2},0\right)\) to line (2) is

\(P_2=\frac{\left|bcosθ(-\sqrt{a^2-b^2},0)+asinθ(0)-ab\right|}{\sqrt{b^2cos^2θ+a^2sin^2θ}}\)

\(=\frac{\left|bcosθ\sqrt{a^2-b^2}+ab\right|}{\sqrt{b^2cos^2θ+a^2sin^2θ }}.....(3)\)

On multiplying equations (2) and (3), we obtain. 

\(P_1P_2=\frac{\left|bcosθ\sqrt{a^2-b^2}-ab\right|\left|bcosθ\sqrt{a^2-b^2}+ab\right|}{\left(\sqrt{b^2cos^2θ+a^2sin^2θ}\right)^2}\)

\(=\frac{\left|bcosθ\sqrt{a^2-b^2}-ab\right|\left|bcosθ\sqrt{a^2-b^2}+ab\right|}{\left(b^2cos^2θ+a^2sin^2θ\right)}\)

\(= \frac{\left|b^2cos^2θ(a^2-b^2)-a^2b^2\right|}{(b^2cos^2θ+a^2sin^2θ)}\)

\(= \frac{\left|a^2b^2cos^2θ-b^4cos^2θ-a^2b^2\right|}{(b^2cos^2θ+a^2sin^2θ)}\)

\(= \frac{b^2\left|a^2cos^2θ-b^2cos^2θ-a^2\right|}{b^2cos^2θ+a^2sin^2θ}\)

\(= \frac{b^2\left|a^2cos^2θ-b^2cos^2θ-a^2sin^2θ-a^2cos^2θ\right|}{b^2cos^2θ+a^2sin^2θ}\)       \([sin^2θ+cos^2θ=1]\)

\(=\frac{ b^2\left|-\left(b^2cos^2θ+a^2sin^2θ\right)\right|}{b^2cos^2θ+a^2sin^2θ}\)

\(= \frac{b^2\left(b^2cos^2θ+a^2sin^2θ\right)}{b^2cos^2θ+a^2sin^2θ}\)

\(= b^2.\)
Hence, proved.