Question

Find the points on the $x-axis$ such that their perpendicular distance from the line $\frac{x}{a}+\frac{y}{b}=1$ is $a,a,b>0$.

Answer 1

Let the point be $(x,o)$

$\therefore$ distance $=\frac{|\frac{x}{a}+\frac{o}{b}-1|}{\sqrt{\frac{1}{a^2+\frac{1}{b^2}}}}$

$\therefore \frac{|\frac{x}{a}-1|}{\sqrt{\frac{1}{a^2+\frac{1}{b^2}}}}=a$

$\therefore {|\frac{x}{a}-1|}^2=a^2(\frac{1}{a^2}+\frac{1}{b^2})$

$\therefore {|\frac{x}{a}-1|}^2=1+\frac{a^2}{b^2}$

$\therefore \frac{x}{a}-1=\sqrt{1+\frac{a^2}{b^2}}$

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

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