# Prove that the product of the lengths of perpendiculars drawn from the points

Question:

Prove that the product of the lengths of perpendiculars drawn from the points

$\mathrm{A}\left(\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ and $\mathrm{B}\left(-\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ to the line $\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1$, is $\mathrm{b}^{2}$

Solution:

Given: Point $\mathrm{A}\left(\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right), \mathrm{B}\left(-\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ and line $\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1$

To Prove: The product of the lengths of perpendiculars drawn from the points

$\mathrm{A}\left(\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ and $\mathrm{B}\left(-\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}, 0\right)$ to the line $\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{b}} \sin \theta=1$, is $\mathrm{b}^{2}$

Formula used:

We know that the length of the perpendicular from $(m, n)$ to the line $a x+b y+c$ $=0$ is given by,

$D=\frac{|a m+b n+c|}{\sqrt{a^{2}+b^{2}}}$

The equation of the line is $\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta-1=0$

At point $A, m=\sqrt{a^{2}-b^{2}}$ and $n=0, a=\frac{\cos \theta}{a} b=\frac{\sin \theta}{b} c=-1$

$D_{1}=\frac{\left|\frac{\cos \theta}{a}\left(\sqrt{a^{2}-b^{2}}\right)+\frac{\sin \theta}{b}(0)-1\right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^{2}+\left(\frac{\sin \theta}{b}\right)^{2}}}$

$D_{1}=\frac{\left|\frac{\cos \theta}{a}\left(\sqrt{a^{2}-b^{2}}\right)-1\right|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}}$

At point $B, m=-\sqrt{a^{2}-b^{2}}$ and $n=0, a=\frac{\cos \theta}{a} b=\frac{\sin \theta}{b} c=-1$

$D_{2}=\frac{\left|\frac{\cos \theta}{a}\left(-\sqrt{a^{2}-b^{2}}\right)+\frac{\sin \theta}{b}(0)-1\right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^{2}+\left(\frac{\sin \theta}{b}\right)^{2}}}$

$D_{2}=\frac{\left|\frac{\cos \theta}{a}\left(-\sqrt{a^{2}-b^{2}}\right)-1\right|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}}=\frac{\left|\frac{\cos \theta}{a}\left(\sqrt{a^{2}-b^{2}}\right)+1\right|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}}$

Product of the lengths of perpendiculars drawn from the points $A$ and $B$ is $D_{1} \times$ $\mathrm{D}_{2}$

$D_{1} \times D_{2}=\frac{\left|\frac{\cos \theta}{a}\left(\sqrt{a^{2}-b^{2}}\right)-1\right|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}} \times \frac{\mid \frac{\cos \theta}{a}\left(\sqrt{a^{2}-b^{2}}\right)+1}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}}=\frac{\left|\frac{\cos ^{2} \theta}{a^{2}}\left(a^{2}-b^{2}\right)-1\right|}{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}$

(In the numerator we have $(x-y) \times(x+y)=x^{2}+y^{2}$ and $\left.\sin ^{2} \theta+\cos ^{2} \theta\right)$

$D_{1} \times D_{2}=\frac{\left|\frac{\cos ^{2} \theta \times a^{2}}{a^{2}}+\frac{\cos ^{2} \theta \times\left(-b^{2}\right)}{a^{2}}-\cos ^{2} \theta-\sin ^{2} \theta\right|}{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}=\frac{\left|\cos ^{2} \theta+\frac{\cos ^{2} \theta \times\left(-b^{2}\right)}{a^{2}}-\cos ^{2} \theta-\sin ^{2} \theta\right|}{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}$

$D_{1} \times D_{2}=\frac{\left|\frac{\cos ^{2} \theta \times\left(-b^{2}\right)}{a^{2}}-\sin ^{2} \theta\right|}{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}=b^{2} \times \frac{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}=b^{2}$

$D_{1} \times D_{2}=b^{2}$

Product of the lengths of perpendiculars drawn from the points $A$ and $B$ is $b^{2}$