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.