Q. A normal is drawn at a $P(x,y)$ of a curve. It meets the $x$-axis at Q. if $PQ$ is of constant length $k$, then the differential equation describing such a curve is
A) $y\ \frac{dy}{dx} = \pm \sqrt{k^{2} - y^{2}}$
B) $x\ \frac{dy}{dx} = \pm \sqrt{k^{2} - x^{2}}$
C) $y\ \frac{dy}{dx} = \pm \sqrt{y^{2} - k^{2}}$
D) $x\ \frac{dy}{dx} = \pm \sqrt{x^{2} - k^{2}}$
Explanation
(a)
Given, $k = PQ =$length of normal
$\Rightarrow$ $k = y\sqrt{1 + \left( \frac{dy}{dx} \right)^{2}}$
$\Rightarrow$ $\frac{k^{2}}{y^{2}} = 1 + \left( \frac{dy}{dx} \right)^{2}$
$\therefore$ $y\ \frac{dy}{dx} = \pm \sqrt{k^{2} - y^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.