JEE Main 2026 — Mathematics Coordinate Geometry

Given the equation of the parabola $y = x^2 + px + q$y=x2+px+q.

Since it passes through $(1, -1)$(1,−1), we have:

$$-1 = 1 + p + q \Rightarrow q = -p - 2$$−1=1+p+q⇒q=−p−2

The vertex of the parabola $y = x^2 + px + q$y=x2+px+q occurs at $x = -\dfrac{p}{2}$x=−2p​.

The $y$y-coordinate of the vertex is:

$$y_v = \left(-\dfrac{p}{2}\right)^2 + p\left(-\dfrac{p}{2}\right) + q = -\dfrac{p^2}{4} + q$$yv​=(−2p​)2+p(−2p​)+q=−4p2​+q

The distance between the vertex and the $x$x-axis is $|y_v|$∣yv​∣. Substituting $q = -p - 2$q=−p−2 into the expression for $y_v$yv​:

$$|y_v| = \left|-\dfrac{p^2}{4} - p - 2\right| = \dfrac{1}{4}|p^2 + 4p + 8| = \dfrac{1}{4}|(p+2)^2 + 4|$$∣yv​∣=​−4p2​−p−2​=41​∣p2+4p+8∣=41​∣(p+2)2+4∣

To minimize this distance, the term $(p+2)^2$(p+2)2 must be minimum. Since $(p+2)^2 \ge 0$(p+2)2≥0 for all real $p$p, the minimum value occurs at $p = -2$p=−2.

Substituting $p = -2$p=−2 into $q = -p - 2$q=−p−2, we get:

$$q = -(-2) - 2 = 0$$q=−(−2)−2=0

Therefore, the value of $p^2 + q^2$p2+q2 is:

$$p^2 + q^2 = (-2)^2 + 0^2 = 4$$p2+q2=(−2)2+02=4

Answer: $4$4