# The shortest distance between the line $\dpi{100} y-x = 1$ and the curve $\dpi{100} x=y^{2}$ is Option 1) $\frac{2\sqrt{3}}{8}$ Option 2) $\frac{3\sqrt{2}}{5}$ Option 3) $\frac{\sqrt{3}}{4}$ Option 4) $\frac{3\sqrt{2}}{8}$

As we learnt in

Perpendicular distance of a point from a line -

$\rho =\frac{\left | ax_{1}+by_{1}+c\right |}{\sqrt{a^{2}+b^{2}}}$

- wherein

$\rho$  is the distance from the line $ax+by+c=0$ .

Let (a2 , a) be a point on x = y2

Distance between (a2, a) and x- y+ 1= is

$\frac{a^{2}-a+1}{\sqrt{2}}= \frac{1}{\sqrt{2}}[(a-\frac{1}{2})^{2}+\frac{3}{4}]$

It is minimum when $a=\frac{1}{2}$

So minimum distance = $\frac{3}{4\sqrt{2}}$

Option 1)

$\frac{2\sqrt{3}}{8}$

This is incorrect.

Option 2)

$\frac{3\sqrt{2}}{5}$

This is incorrect.

Option 3)

$\frac{\sqrt{3}}{4}$

This is incorrect.

Option 4)

$\frac{3\sqrt{2}}{8}$

This is incorrect.

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