AOD

find the pint on the curve y^2=4x which is nearest to the point (2, -8)

1357
Manish Shankar ·

Let (x1, y1) be the nearest point on the curve to the given point.

So there are two methods

First one: Find the distance between the two point and minimize it.

Second: The point will be nearest if normal at (x1, y1) passes through given point (Why?)

1708
man111 singh ·

\hspace{-16}$Using Manish Sir hint......\\\\\\First we will Calculate the parametric coordinate of$\bf{y^2=4x}$\\\\\\ which is$\bf{x=t^2}$and$\bf{y=2t}$\\\\\\ So Coordibate of point$\bf{P(t^2,2t)}$and Let$\bf{Q(2,-8)}$\\\\\\ Now Distance b/w$\bf{PQ = \sqrt{(t^2-2)^2+(2t+8)^2}}$\\\\\\ Now we have to Minimize$\bf{PQ}$\\\\\\ Now Let$\bf{f(t) = (PQ)^2 = (t^2-2)^2+4(t+4)^2}$\\\\\\ bcz Minimizing$\bf{PQ}$means Minimize$\bf{(PQ)^2}$\\\\\\ So Using First Derivative Test.....\\\\\\$\bf{f^{'}(t) = 2(t^2-2)\cdot 2t+4\cdot 2 \cdot (t+4)}$\\\\\\ \hspace{-16}$For Max. and Min. put $\bf{f^{'}(t)=0\Rightarrow t^3-2t+2t+8=0}$\\\\\\ we Get $\bf{t^3+2^3=0\Rightarrow t = -2}$ and $\bf{2}$ Complex values.\\\\\\ Now $\bf{f^{''}(t)=......,}$ which is $\bf{>0}$ for $\bf{t=-2}$\\\\\\ So $\bf{t=-2}$ is a point of Minimum.\\\\\\ So $\bf{f(-2)=(PQ)^2 = 20\Rightarrow PQ = 4\sqrt{5}}$\\\\\\ Which is occur at $\bf{(x,y) = (4,-4)}$