Question

Transform the equation $\frac{x}{a}+\frac{y}{b}=1$ into normal form. If the perpendicular distance of the straight line from origin is $p$,deduce that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

$\frac{x}{a}+\frac{y}{b}=1$ .....$\left(1\right)$ where $a>0,b>0$

$\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}=\sqrt{\frac{a^2+b^2}{a^2{b}^2}}=\frac{\sqrt{a^2+b^2}}{ab}$

Divide $\left(1\right)$ through out by $\frac{\sqrt{a^2+b^2}}{ab}$

$\Rightarrow \frac{x}{a\frac{\sqrt{a^2+b^2}}{ab}}+\frac{y}{b\frac{\sqrt{a^2+b^2}}{ab}}=\frac{1}{\frac{\sqrt{a^2+b^2}}{ab}}$

$\Rightarrow x\left(\frac{b}{\sqrt{a^2+b^2}}\right)+y\left(\frac{a}{\sqrt{a^2+b^2}}\right)=\frac{ab}{a^2+b^2}$ .......$\left(1\right)$

Slope of $\left(1\right)$ is given by

$\tan \alpha =-\frac{\frac{1}{a}}{\frac{1}{b}}=\frac{-b}{a}$ which is negative, hence $\alpha$ is obtuse.

$\tan \alpha =\frac{-b}{a}=\frac{perpendicular}{base}$

Hypotenuse$=\sqrt{{\left(base\right)}^2+{\left(perpendicular\right)}^2}=\sqrt{a^2+b^2}$

For obtuse angles $\sin$ is positive and $\cos$ is negative

$\sin \alpha =\frac{b}{\sqrt{a^2+b^2}}$ and $\cos \alpha =\frac{-b}{\sqrt{a^2+b^2}}$

Hence, $\left(1\right)$ becomes

$x(-\cos \alpha )+y\left(\sin \alpha \right)=\frac{ab}{\sqrt{a^2+b^2}}$

We know that $\sin A=\sin (\pi -A)$ and $\cos A=-\cos (\pi -A)$

$\Rightarrow x\cos (\pi -\alpha )+y\sin (\pi -\alpha )=\frac{-ab}{\sqrt{a^2+b^2}}$ which is the required normal form where $\tan \alpha$ is slope of given line.

hence right hand side must give perpendicular distance of the straight line from the origin.

$\Rightarrow p=\frac{ab}{\sqrt{a^2+b^2}}$

$\Rightarrow p^2=\frac{a^2{b}^2}{a^2+b^2}$

$\Rightarrow \frac{1}{p^2}=\frac{a^2+b^2}{a^2{b}^2}$

$\Rightarrow \frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

hence proved.