JEE Main 2026 — Apr 2, Shift 2 — Mathematics solution

JEE Main2026Apr 2, Shift 2Mathematics

Let the parabola $y = x^{2} + p x + q$ 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 :

A
$2$
B
$4$
C
$5$
D
$8$

Solution

The point $(1, - 1)$ on the parabola gives $p + q = - 2$ , so $q = - 2 - p$ .

The vertex's $y$ -coordinate is $\frac{4 q - p ^{2}}{4} = \frac{4 ( - 2 - p ) - p ^{2}}{4} = - \frac{( p + 2 ) ^{2} + 4}{4}$ .

The distance from the $x$ -axis is the absolute value of this expression. Since $(p + 2)^{2} + 4 \geq 4$ , the minimum distance occurs when $(p + 2)^{2} = 0$ , i.e. $p = - 2$ .

Then $q = 0$ , so $p^{2} + q^{2} = 4$ .

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