JEE Main 2026 · 2 Apr · Shift 2 · Mathematics Q12

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MathematicsSingle correct+4 / -1Question 12 of 75

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 :

Answer and worked solution

Official answer: B

Through (1, −1): $q=-2-p$ .

Vertex height. $y_v=q-\frac{p^2}4=-\left[\frac{(p+2)^2}4+1\right]$ . Its distance from the x-axis, $\frac{(p+2)^2}4+1$ , is least at $p=-2$ , which gives $q=0$ .

Value. $p^2+q^2=4$ .

(B)

Topics:Quadratic expressions Parabola

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JEE Main 2026 · 2 Apr · Shift 2, Mathematics question 12