The parabola $y=x^2+px+q$ cuts the straight line $y=2x-3$ at a point with abscissa 1. If the distance between the vertex of parabola and the X-axis is least, then 

$$\begin{gathered} \text{ We have, }y=x^2+px+q \\ \text{ Cuts the straight line }y=(2x-3)\text{, where }x=1\text{ i.e. }y=-1 \end{gathered}$$
$$\begin{gathered} \Rightarrow (1,-1)\text{ lies on }y=x^2+px+q \\ \Rightarrow -1=1+p+q\Rightarrow p+q=-2 \\ \text{ Vertex of parabola is }\left(-\frac{p}{2},\frac{-\left(p^2-4q\right)}{4}\right)=\left(\frac{-p}{2},\frac{-p^2+4q}{4}\right) \\ \text{ Distance between vertex and }\mathrm{X}\text{-axis } \\ \begin{array}{l}\frac{\left|p^2-4q\right|}{4}=\frac{1}{4}\left|p^2-4(-2-p)\right|=\frac{1}{4}\left|p^2+4p-8\right|\\ =\frac{p^2+4p+8}{4}\text{ as }{p}^2+4p+8>0 \\ \text{ Let }L=\frac{1}{4}\left(p^2+4p+8\right),\frac{dL}{dp}=\frac{1}{4}(2p+4)=\frac{p}{2}+1 \\ \text{ For least value of }L,\frac{dL}{dp}=0\Rightarrow \frac{p}{2}+1=0\Rightarrow p=-2 \\ \text{ Also, }p+q=-2\Rightarrow q=-2-p\Rightarrow q=-2-(-2)=0 \\ \therefore p=-2\text{ and }q=0 \\ \text{ And least distance }=\frac{1}{4}(4-8+8)=1\end{array} \end{gathered}$$