Question

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

A
325
B
34
C
328
D
238

Solution

## The correct option is C $$\cfrac { 3\sqrt { 2 } }{ 8 }$$let the point on parabola be $$(t,t^2)$$ distance of point to line $$y-x-1=0$$ is $$p=\dfrac{t^2-t+1}{\sqrt{(1)^2+(-1)^2}}=\dfrac{t^2-t+1}{\sqrt2}$$given p is minimumso $$\dfrac{dp}{dt}=2t-1=0$$$$=>t=\dfrac{1}{2}$$now $$\dfrac{d^2p}{dt^2}=2>0$$so pi minimum at $$t=\dfrac{1}{2}$$therefore $$p=\dfrac{(\dfrac{1}{2})^2-\dfrac{1}{2}+1}{\sqrt2}=\dfrac{3\sqrt2}{8}$$Maths

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