Question

What are the points on X-axis whose perpendicular distance from the straight line \[\frac{x}{a} + \frac{y}{b} = 1\] is a ?

Answer 1

Let (t, 0) be a point on the x-axis.
It is given that the perpendicular distance of the line \[\frac{x}{a} + \frac{y}{b} = 1\] from a point is a.

\[\therefore \left| \frac{\frac{t}{a} + 0 - 1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right| = a\]

\[ \Rightarrow a^2 \left( \frac{1}{a^2} + \frac{1}{b^2} \right) = \frac{t^2}{a^2} + 1 - \frac{2t}{a}\]

\[ \Rightarrow 1 + \frac{a^2}{b^2} = \frac{t^2}{a^2} + 1 - \frac{2t}{a}\]

\[ \Rightarrow \frac{a^2}{b^2} = \frac{t^2}{a^2} - \frac{2t}{a}\]

\[\Rightarrow b^2 t^2 - 2a b^2 t - a^4 = 0\]

\[ \Rightarrow t = \frac{2a b^2 \pm 2\sqrt{a^2 b^4 + b^2 a^4}}{2 b^2}\]

\[ \Rightarrow t = \frac{a}{b}\left( b \pm \sqrt{a^2 + b^2} \right)\]

Hence, the required points on the x-axis are

\[\left( \frac{a}{b}\left( b - \sqrt{a^2 + b^2} \right), 0 \right) \text { and } \left( \frac{a}{b}\left( b + \sqrt{a^2 + b^2} \right), 0 \right)\].