JEE Main 2026 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2026, Shift 2, Maths Q12

Question 12 of 75 in this shift, Maths question 12 of 25, Section A.

Let the parabola

y = x^2 + px + q

passing through the point

(1, -1)

be such that the distance between its vertex and the

x

-axis is minimum. Then the value of

p^2 + q^2

is:

(1) $2$
(2) $\frac{4}{4}$ Official answer
(3) $5$
(4) $8$

Official answer

Option 2

NTA final key.

Chapter: Conic Sections
Topic: Quadratic Roots
Idea tested: Vertex and Extrema Relation

Same topic in other shifts

All Quadratic Roots questions

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5 Apr 2026, Shift 2 · Q1Let $\alpha, \beta$ be the roots of the equation $x^2-x+p=0$ and $\gamma, \delta$ be the roots the equation $x^2-4x+q=0$ ;…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 42 of the 174 JEE Main shifts held from 2017 to 2026.