The shortest distance between the line $\mathrm{y}-\mathrm{x}=1$ and the curve $\mathrm{x}=\mathrm{y}^2$ is

$$ \begin{aligned} & x-y+1=0 \\ & x=y^2 \\ & 1=2 y \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{1}{2 y}=\text { Slope of given line (1) } \\ & \frac{1}{2 y}=1 \Rightarrow y=\frac{1}{2} \Rightarrow y=\frac{1}{2} \Rightarrow x=\left(\frac{1}{2}\right)^2=\frac{1}{4} \Rightarrow(x, y)=\left(\frac{1}{4}, \frac{1}{2}\right) \end{aligned} $$ $\therefore$ The shortest distance is $\frac{\left|\frac{1}{4}-\frac{1}{2}+1\right|}{\sqrt{1+1}}=\frac{3}{4 \sqrt{2}}=\frac{3 \sqrt{2}}{8}$ Directions: Question number 86 to 90 are Assertion - Reason type questions. Each of these questions contains two statements Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice