Question

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

Answer 1

Perpendicular Length from Point to line $P(x_1,y_1)$

$ax+by+c=0$

$d=\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

given line $\frac{x}{a}cos\theta +\frac{y}{b}sin\theta =1$

given Point $(-\sqrt{a^2-b^2},0)$ and $(\sqrt{a^2-b^2},0)$

Product of Perpendicular distance from

both Point to line

$=\frac{|-\frac{\sqrt{a^2-b^2}}{a}cos\theta +0-1|}{\sqrt{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2\theta }{b^2}}}\times \frac{|\sqrt{\frac{a^2-b^2}{a}}cos\theta +0-1|}{\sqrt{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2}{b^2}}}$

$=\frac{\frac{-co{s}^2\theta }{a^2}(a^2-b^2)+1}{\frac{co{s}^2\theta }{a^2}+\frac{si{n}^2\theta }{b^2}}=\frac{(-co{s}^2\theta +\frac{b^2}{a^2}co{s}^2\theta +1)a^2{b}^2}{(b^{2}co{s}^2\theta +a^{2}si{n}^2\theta )}$

$=\frac{\frac{(a^{2}si{n}^2\theta +b^{2}co{s}^2\theta )a^2{b}^2}{a^2}}{(a^{2}si{n}^2\theta +b^{2}co{s}^2\theta )}=b^2$