JEE PYQ: Application Of Derivatives Question 50
Question 50 - 2020 (08 Jan Shift 2)
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:
(1) $\sqrt{2}$
(2) $4\sqrt{2}$
(3) $2$
(4) $2\sqrt{2}$
Show Answer
Answer: (4)
Solution
$\frac{dy}{dx} = \frac{x+y}{3y-x}$. At $(2,2)$: slope $= \frac{4}{4} = 1$. Normal: $y - 2 = -(x-2) \Rightarrow x + y = 4$. Distance from origin $= \frac{4}{\sqrt{2}} = 2\sqrt{2}$.
BETA
