The length of the perpendicular from the origin,

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:

  1. (1) $\sqrt{2}$

  2. (2) $4 \sqrt{2}$

  3. (3) 2

  4. (4) $2 \sqrt{2}$


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

$=\left|\frac{0+0-4}{\sqrt{2}}\right|=2 \sqrt{2}$

 

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