JEE Main 2020 (Online) 8th January Evening ShiftQuestion 1
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 :
Differentiation and its Application
Solution
Concept. The curve is the pair of lines $x+3y=0$ and $x-y=0$; $(2,2)$ is on $y=x$. Formulas used. (1) $x^2+2xy-3y^2=(x+3y)(x-y)$(2) Distance of $(0,0)$ from $px+qy+r=0$: $\dfrac{|r|}{\sqrt{p^2+q^2}}$Step 1 - Tangent. By (1), near $(2,2)$ the curve is the line $y=x$, slope $1$. Step 2 - Normal. Slope $-1$ through $(2,2)$: $x+y=4$. Step 3 - Distance. By (2), $\frac{4}{\sqrt2}=2\sqrt2$. Answer: (C)