Question

Prove that the product the lengths of perpendicular form $P(x_1,y_1)$ to the line represented ${ax}^2+2hxy+{by}^2=0$ is $\frac{{ax}_1^2+2{hx}_1{y}_1+{by}_1^2}{\sqrt{{(a-b)}^2+{4h}^2}}$

Answer 1

Let $m_1$ and $m_2$ be the slopes of the lines representyed by

$a{x}^2+2hxy+b{y}^2=0$

$\therefore m_1+m_2=-\frac{2h}{b}$ and $m_1{m}_2=\frac{a}{b}...\left(1\right)$

The separate equation of the lines represented by $a{x}^2+2hxy+b{y}^2=0$ are

$y=m_{1}x$ and $y=m_{2}x$

i.e. $m_{1}x-y=0$ and $m_{2}x-y=0$

length of perpendicular from $P(x_1,y_1)$ on

$m_{1}x-y=0$ is $\left|\frac{m_1{x}_1-y_1}{\sqrt{m_1^2+1}}\right|$

Length of perpendicular from $P(x_1,y_1)$ on

$m_{2}x-y=0$ is $\left|\frac{m_2{x}_1-y_1}{\sqrt{m_2^2+1}}\right|$

$\therefore$ product of lengths of perpendiculars

$\left|\frac{m_1{x}_1-y_1}{\sqrt{m_1^2+1}}\right|\times \left|\frac{m_2{x}_1-y_1}{\sqrt{m_2^2+1}}\right|$

$=\left|\frac{m_1{m}_2{x}_1^2-(m_1+m_2)x_1{y}_1+y_1^2}{\sqrt{m_1^2{m}_2^2+m_1^2+m_2^2+1}}\right|$

$\frac{m_1{m}_2{x}_1^2-(m_1+m_2)x_1{y}_1+y_1^2}{\sqrt{m_1^2{m}_2^2+(m_1+m_2{)}^2-2m_1{m}_2+1}}$

$=\left|\frac{\frac{a}{b}.x_1^2-\frac{-2h}{b}{x}_1{y}_1+y_1^2}{\sqrt{\frac{a^2}{b^2}+\frac{4h^2}{b^2}-\frac{2a}{b}}+1}\right|...\left(By\right(1\left)\right)$

$=\left|\frac{a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2}{\sqrt{a^2+4h^2-2ab+b^2}}\right|$

$=\left|\frac{a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2}{\sqrt{(a^2-2ab+b^2)+4h^2}}\right|$

$=\left|\frac{a{x}_1^2+2h{x}_1{y}_1+b{y}_1^2}{\sqrt{(a^2-b^2{)}^2+4h^2}}\right|$