JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Application of Derivatives

The point $(1,-1)$ lies on the parabola, so $-1=1+p+q$, giving $q=-2-p$. The vertex of $y=x^2+px+q$ has $y$-coordinate $q-\frac{p^2}{4}$, and its distance from the x-axis is $\left|q-\frac{p^2}{4}\right|$. Substituting $q=-2-p$,

which is always negative, so the distance equals $\frac{(p+2{)}^2+4}{4}$. This is least when $(p+2{)}^2=0$, i.e. $p=-2$, and then $q=0$. Hence $p^2+q^2=4+0=4$.