Question

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

A
832
B
43
C
34
D
328

Solution

## The correct option is A $$\dfrac{8}{3\sqrt{2}}$$$$y^2 = x \\ \Rightarrow 2y = \dfrac{dx}{dy} \, [ \textrm{Differentiating both sides w.r.t. 'y' } ] \\ \left[ - \dfrac{dx}{dy} \right]_{x_1, y_1} = -2y_1$$$$\therefore$$ Slope of the normal of the parabola $$y^x$$ at $$P ( x_1, y_1) = -2y_1$$Again slope of the line y - x = 1 is 1.$$\therefore (-2y_1).(1) = -1 \\ \Rightarrow y_1 = \dfrac{1}{2}$$Again $$P (x_1, y_1)$$ lie on $$y^2 = x$$$$\therefore x_1 = y_1^2 = \dfrac{1}{4}.$$$$\therefore$$ The point P is $$\left( \dfrac{1}{4} , \dfrac{1}{2} \right)$$$$\therefore$$ The distance from $$\left( \dfrac{1}{4} , \dfrac{1}{2} \right)$$ to y - x = 1$$d = \dfrac{| \frac{1}{2} - \frac{1}{4} - 1 | } { \sqrt{ (1)^2 + (-1) ^2 }} = \dfrac{ |-\frac{1}{2} - \frac{1}{4} | } {\sqrt2} \\ = \dfrac{3}{4\sqrt2} = \dfrac{3\sqrt2}{8}$$Mathematics

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