JEE Main 2020 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 8 January 2020, Shift 2, Maths Q60

Question 60 of 75 in this shift, Maths question 10 of 25, Section A.

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) $2$
(3) $2\sqrt{2}$ Official answer
(4) $4\sqrt{2}$

Official answer

Option 3

NTA final key (Jan 2020).

Chapter: Application of Derivatives
Topic: Tangents and Normals
Idea tested: Equation of Normal

Same idea in other shifts

Asked 6× in all

4 Aug 2021, Shift 1 · Q68The equation of the normal to the curve $y=e^{-|x|}$ at a point, where the curve cuts the line $x=-1$ isEasySingle correct
26 Feb 2021, Shift 2 · Q83Let the normals at all the points on a given curve pass through a fixed point $(a, b)$ . If the curve passes through $(3, -3)$ and…MediumNumerical value
2 Sep 2020, Shift 1 · Q64Let P(h, k) be a point on the curve $y=x^{2}+7x+2$ , nearest to the line, $y=3x-3$ . Then the equation of the normal to the curve at P is :MediumSingle correct
2 Sep 2020, Shift 2 · Q60The equation of the normal to the curve $y=(1+x)^{2y}+\cos^2(\sin^{-1}x)$ at $x=0$ is :MediumSingle correct
2 Apr 2017 (offline) · Q72The normal to the curve $y(x-2)(x-3)=x+6$ at the point where the curve intersects the $y$ -axis passes through the point :MediumSingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.