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  4. The length of the perpendicular from the origin, on the normal to the curve, x2+2xy-3y2=0 at the point (2, 2) is :

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Q. 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 :

JEE MainJEE Main 2020Application of Derivatives

Solution:

$ x^{2}+2xy-3y^{2}=0$
$m_{N} =$ slope of normal drawn to curve at $\left(2,2\right)$ is $-1$
$L : x + y = 4$.
perpendicular distance of L from $\left(0,0\right)$
$= \frac{\left|0+0-4\right|}{\sqrt{2}} = 2\sqrt{2}$