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  • Find the point on the curve which is nearest to the point (2, 1).  

Find the point on the curve y^{2} = 4x which is nearest to the point (2, 1).

 

Answers (1)

$ Let $ (x,y) $ on the $ y^{2} = 4x $ is nearest point to the (2,1)

\\ $ Distance D = $ \sqrt{(x-2)^2+ (y-1)^2} \\ D = \sqrt{(\frac{y^2}{4}{}-2)^2+ (y-1)^2} \\ D = \sqrt{(y^2-8)^2+ 16(y-1)^2}/4 \\ \frac{dD}{dy} = 0 \\\\ \frac{2(y^2-8)(2y)+32(y-1)}{8\sqrt{(y^2-8)^2+ 16(y-1)^2}/4} = 0 \\\\ y^3-8y + 8y-8 = 0 \\\\ y = 2 \Rightarrow x =1 \\

\\\frac{d^2D}{dy^2}_{y = 2}>0

y=2 has minima 

\\ $ Shortest Distance D = $ \sqrt{(1-2)^2+ (2-1)^2} = \sqrt{2} $ unit \\

 

Posted by

Safeer PP

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