JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 24 June 2022, Shift 2, Maths Q19

Question 19 of 90 in this shift, Maths question 19 of 30, Section A.

The slope of normal at any point

(x, y)

x > 0

y > 0

on the curve

y = y(x)

is given by

\frac{x^2}{xy - x^2y^2 - 1}

. If the curve passes through the point

(1, 1)

, then

e \cdot y(e)

is equal to

(1) $\frac{1-\tan(1)}{1+\tan(1)}$
(2) $\tan(1)$
(3) $1$
(4) $\frac{1+\tan(1)}{1-\tan(1)}$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Differential Equations
Topic: Variable Separable Form
Idea tested: Geometric Applications

Same idea in other shifts

Asked 3× in all

25 Jul 2022, Shift 2 · Q11Let a smooth curve $y=f(x)$ be such that the slope of the tangent at any point $(x,y)$ on it is directly proportional to…MediumSingle correct
24 Jun 2022, Shift 2 · Q8A particle is moving in the $xy$ -plane along a curve $C$ passing through the point $(3, 3)$ . The tangent to the curve $C$ at the point $P$ …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.