Q53 / 90 · JEE Main 2011 · Mathematics

The shortest distance between line $y-x=1$ and curve $x=y^2$ is

  • Correct answer
    A
    $\frac{3 \sqrt{2}}{8}$
  • B
    $\frac{8}{3 \sqrt{2}}$
  • C
    $\frac{4}{\sqrt{3}}$
  • D
    $\frac{\sqrt{3}}{4}$

Solution

$$ P=\left(y^2, y\right) $$ Perpendicular distance from $P$ to $x-y+1=0$ is $\frac{\left|y^2-y+1\right|}{\sqrt{2}}$ $$ \begin{aligned} & \mathrm{y}^2-\mathrm{y}+1>0 \forall \mathrm{y} \in \mathrm{R} \\ & \therefore \text { Coefficient } \mathrm{y}^2>0 \\ & \therefore \text { Min value }=\frac{1}{\sqrt{2}}\left(\frac{4 \mathrm{ac}-\mathrm{b}^2}{4 \mathrm{a}}\right)=\frac{3}{4 \sqrt{2}} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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