JEE Main 2026 — Parabola Question with Solution

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

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

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

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

The $y$-coordinate of the vertex is:

$y_v={\left(-\frac{p}{2}\right)}^2+p\left(-\frac{p}{2}\right)+q=-\frac{p^2}{4}+q$

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

$|y_v|=\left|-\frac{p^2}{4}-p-2\right|=\frac{1}{4}|p^2+4p+8|=\frac{1}{4}|(p+2{)}^2+4|$

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

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

$q=-(-2)-2=0$

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

$p^2+q^2=(-2{)}^2+0^2=4$

Answer: $4$