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Question

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


A
325
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B
34
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C
328
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D
238
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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 minimum
so $$\dfrac{dp}{dt}=2t-1=0$$
$$=>t=\dfrac{1}{2}$$
now $$\dfrac{d^2p}{dt^2}=2>0$$
so pi minimum at $$t=\dfrac{1}{2}$$
therefore $$p=\dfrac{(\dfrac{1}{2})^2-\dfrac{1}{2}+1}{\sqrt2}=\dfrac{3\sqrt2}{8}$$

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