Let the parabola y = x^2 + px + q passing through the point (1, -1) be such that the dista…

Question

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 :

Accepted Answer

parabola passes through $(1,-1)$ so

$-1=1+p+q$

$p+q=-2$

$q=-2-p\, \, \, \ldots (1)$

Distance from x-axis

$\frac{-D}{4a}=\frac{-({p}^{2}-4q)}{4(1)}=\frac{4q-{p}^{2}}{4}$

$=\frac{4(-2-p)-{p}^{2}}{4}\,$

$=\frac{-8-4p-{p}^{2}}{4}=\frac{-({p}^{2}+4p+8)}{4}=\frac{-((p+2{)}^{2}+4)}{4}$

Minimum at $p=-2	o q=0$

${p}^{2}+{q}^{2}=4+0=4$

Solutions · 1