Question

Transform the equation $\frac{x}{a}+\frac{y}{b}=1$ into normal form when $a>0$ and $b>0$. If the perpendicular distance of stright line form the origin is $p$, deduce that $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

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

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

Divide (i) throughout by $\frac{\sqrt{a^2+b^2}}{ab}$

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

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

Slope of (ii) is given by

$\tan \alpha =-\frac{Coefficientofx}{Coefficientofy}$

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

$\tan \alpha =-\frac{b}{a}=\frac{Perpendicular}{Base}$

$Hypotenuse=\sqrt{Bas{e}^2+Perpendicula{r}^2}=\sqrt{a^2+b^2}$

For obtuse angles sin is positive and cos is negative

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

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

Hence (ii) will become

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

We know that

$\sin \left(A\right)=\sin (\pi -A)$

and $\cos \left(A\right)=-\cos (\pi -A)$

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

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 orogin.

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

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

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

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

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