Question

The length of the perpendicular from the origin, on the normal to the curve, $x^2+2xy-3y^2=0$ at the point $(2,2)$ is:

• A. $2$
• B. $2\sqrt{2}$
• C. $4\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

The correct option is B

$2\sqrt{2}$

Find the length of the perpendicular from the origin based on given information

Given curve, $x^2+2xy-3y^2=0$

Differentiating with respect to $x$

$2x+2y+2x\frac{dy}{dx}-6y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-\left(2x+2y\right)}{\left(2x-6y\right)}$

At $(2,2)$ , $\frac{dy}{dx}=\frac{-(4+4)}{\left(4-12\right)}$

$\frac{dy}{dx}=1$

Hence, the slope of the tangent is $1$

The normal is perpendicular to the tangent,

$\therefore$slope of normal $m=-1$

Hence , equation of the normal at $(2,2)$ is given as

$y-2=-1(x-2)$

$\Rightarrow$ $x+y-4=0$

Perpendicular distance from $(0,0)$ to the normal is given by

$ON=\frac{\left|0+0-4\right|}{\sqrt{1^2+1^2}}$

$\therefore$ $ON=\frac{4}{\sqrt{2}}=\frac{4\sqrt{2}}{\sqrt{2}\times \sqrt{2}}$

$\therefore$ $ON=2\sqrt{2}$

So, the length of the perpendicular from the origin to the normal to curve at $(2,2)$ is $2\sqrt{2}$.

Hence option $\left(B\right)$ is the correct answer.