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Mathematics · Parabola

JEE Main 2026 — 2 April, Evening Shift — Question 34

Let the parabola y=x2+px+q\mathbf{y} = \mathbf{x}^{2} + \mathbf{px} + \mathbf{q} passing through the point (1,−1)(1, -1) be such that the distance between its vertex and the x-axis is minimum. Then the value of p2+q2\mathbf{p}^{2} + \mathbf{q}^{2} is:

  1. Option A:

    2

  2. Option B:

    4

    Correct
  3. Option C:

    5

  4. Option D:

    8

Answer: B

Step-by-step solution

parabola passes through (1,−1)(1,-1) so −1=1+p+q-1=1+\mathrm{p}+\mathrm{q} p+q=−2\mathrm{p}+\mathrm{q}=-2

\mathrm{q}=-2-\mathrm{p} \end{gathered}$$ Distance from x -axis $\frac{-\mathrm{D}}{4 \mathrm{a}}=\frac{-\left(\mathrm{p}^{2}-4 \mathrm{q}\right)}{4(1)}=\frac{4 \mathrm{q}-\mathrm{p}^{2}}{4}$ $=\frac{4(-2-p)-p^{2}}{4} \quad$ from (1) $=\frac{-8-4 \mathrm{p}-\mathrm{p}^{2}}{4}=\frac{-\left(\mathrm{p}^{2}+4 \mathrm{p}+8\right)}{4}=\frac{-\left((\mathrm{p}+2)^{2}+4\right)}{4}$ Minimum at $\mathrm{p}=-2 \Rightarrow \mathrm{q}=0$ $\mathrm{p}^{2}+\mathrm{q}^{2}=4+0=4$

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Evening Shift
Subject
Mathematics
Chapter
Parabola
Topic
Introduction to Parabola
Let the parabola y = x 2 + px + q passing through the point (1, -1)… | JEE Main 2026 PYQ with Solution · DhiX AI