Question:
The length of the perpendicular from the origin, on the normal to the curve, $x^{2}+2 x y-3 y^{2}=0$ at the point $(2,2)$ is:
Correct Option: , 4
Solution:
Given equation of curve is
$x^{2}+2 x y-3 y^{2}=0$
$\Rightarrow 2 x+2 y+2 x y^{\prime}-6 y y^{\prime}=0$
$\Rightarrow x+y+x y^{\prime}-3 y y^{\prime}=0$
$\Rightarrow y^{\prime}(x-3 y)=-(x+y)$
$\Rightarrow \frac{d y}{d x}=\frac{x+y}{3 y-x}$
Slope of normal $=\frac{-d x}{d y}=\frac{x-3 y}{x-3 y}$
Normal at point $(2,2)=\frac{2-6}{2+2}=-1$
Equation of normal to curve $=y-2=-1(x-2)$
$\Rightarrow x+y=4$
$\therefore \quad$ Perpendicular distance from origin
$=\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}$