JEE Main 2025 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 3 April 2025, Shift 2, Maths Q17

Question 17 of 75 in this shift, Maths question 17 of 25, Section A.

The shortest distance between the curves

y^2 = 8x

and

x^2 + y^2 + 12y + 35 = 0

is:

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

Official answer

Option 1

NTA final key.

Chapter: Application of Derivatives
Topic: Maxima and Minima
Idea tested: Shortest Distance Between Curves

Same topic in other shifts

All Maxima and Minima questions

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4 Apr 2026, Shift 2 · Q12 $\max_{0\le x\le\pi}\left(16\sin\left(\frac x2\right)\cos^3\left(\frac x2\right)\right)$ is equal to:MediumSingle correct
5 Apr 2026, Shift 1 · Q12The sum of all the integral values of $p$ such that the equation $3\sin^2 x + 12\cos x - 3 = p$ , $x \in \mathbb{R}$ , has at least one…MediumSingle correct
2 Apr 2025, Shift 1 · Q8If the function $f(x)=2x^3-9ax^2+12a^2x+1$ , where $a>0$ , attains its local maximum and local minimum values at p and q, respectively, such…MediumSingle correct
4 Apr 2025, Shift 2 · Q16Let $f$ be a differentiable function on $\mathbf{R}$ such that $f(2)=1$ , $f'(2)=4$ . Let…HardSingle 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 42 of the 174 JEE Main shifts held from 2017 to 2026.