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

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