The parabola y = x² + px + q cuts the straight line y = 2x – 3 at a point with abscissa 1. Then the values of p and q for which the distance between the vertex of the parabola and the x-axis is the minimum is

$\text{ The parabola }\mathrm{y}={\mathrm{x}}^2+\mathrm{px}+\mathrm{q}\text{ cuts the straight line }\mathrm{y}=2\mathrm{x}-3\text{ at a point with abscissa }1\text{. }$
Therefore, y = 2 – 3 = –1.
So, the point is (1, –1). This point lies on the parabola.
Therefore, –1 = 1 + p + q or p + q = –2 ….(i)Distance between vertex and x-axis
$=\frac{{\mathrm{p}}^2}{4}-\mathrm{q}=\frac{{\mathrm{p}}^2}{4}+\mathrm{p}+2=\frac{1}{4}\left\{(\mathrm{p}+2{)}^2+4\right\}\text{ For minimum, }\mathrm{p}=-2.\text{ Hence, }\mathrm{q}=0$