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Question

Find the shortest distance between the line $$x-y+1=0$$ and the curve $${y}^{2}=x$$.


Solution

Line is $$x-y+1=0$$ and curve $$y^2=x$$

On the line and curve you can take any point that should satisfy the them.

So for line if $$x=t-1$$ and $$y=t$$, $$A(t-1,t)$$ then they are satisfying the line.

For curve $$x=t^2$$ , $$y=t$$, they are satisfying the curve $$B(t^2,t)$$

So using distance formula, we have

$$AB=\sqrt{(t^2-(t-1))^2+(t-t)^2}$$

$$=t^2-t+1$$

Let $$f(t)=t^2-t+1$$

$$f'(t)=2t-1$$

For critical point ,$$f'(t)=0$$

So,

$$2t-1=0$$

$$t=\dfrac{1}{2}$$

For minima ,$$f''(t)>0$$

So, $$f''(t)=2>0$$

So at $$t=\dfrac{1}{2}$$, minima occurs

Hence the minimum distance $$=\dfrac{1}{4}-\dfrac{1}{2}+1=\dfrac{3}{4}$$

Mathematics

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