JEE Main 2009 — Application of Derivatives Question with Solution

$$\begin{array}{rl}& x-y+1=0\\ & x=y^2\\ & 1=2y\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{1}{2y}=\text{ Slope of given line (1) }\\ & \frac{1}{2y}=1\Rightarrow y=\frac{1}{2}\Rightarrow y=\frac{1}{2}\Rightarrow x={\left(\frac{1}{2}\right)}^2=\frac{1}{4}\Rightarrow (x,y)=\left(\frac{1}{4},\frac{1}{2}\right)\end{array}$$ $\therefore$ The shortest distance is $\frac{|\frac{1}{4}-\frac{1}{2}+1|}{\sqrt{1+1}}=\frac{3}{4\sqrt{2}}=\frac{3\sqrt{2}}{8}$ 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