The slope of tangent at any point $(x,y)$ on a curve $y=y(x)$ is $\frac{x^2+y^2}{2xy}$, $x>0$. If $y(2)=0$, then a value of $y(8)$ is

JEE Main 2023 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 10 April 2023, Shift 1, Maths Q10

Question 10 of 90 in this shift, Maths question 10 of 30, Section A.

The slope of tangent at any point

(x,y)

on a curve

y=y(x)

\frac{x^2+y^2}{2xy}

x>0

. If

y(2)=0

, then a value of

y(8)

(1) $4\sqrt{3}$ Official answer
(2) $-4\sqrt{2}$
(3) $2\sqrt{3}$
(4) $-2\sqrt{3}$

Official answer

Option 1

NTA final key.

Chapter: Differential Equations
Topic: Homogeneous Equations
Idea tested: Substitution y = vx

Same idea in other shifts

Asked 10× in all

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18 Mar 2021, Shift 2 · Q86Let $y=y(x)$ be the solution of the differential equation $x\,dy-y\,dx=\sqrt{(x^2-y^2)}\,dx$ , $x\ge1$ , with $y(1)=0$ . If the area bounded…HardNumerical value
9 Jan 2020, Shift 2 · Q64If $\frac{dy}{dx}=\frac{xy}{x^2+y^2}$ ; $y(1)=1$ ; then a value of $x$ satisfying $y(x)=e$ is :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.