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Q.
The nearest point on the curve $y^{2}=4 x$ to $(2,1)$ is
Solution:
Let $P(x, y)$ be any point on the curve $y^{2}=4 x$.
$S=\sqrt{(x-2)^{2}+(y-1)^{2}}$
$\Rightarrow S^{2}=(x-2)^{2}+(y-1)^{2}=\left[\frac{y^{2}}{4}-2\right]^{2}+(y-1)^{2}=f(y)$ (say)
$\Rightarrow f'(y)=2\left[\frac{y^{2}}{4}-2\right]\left[\frac{2 y}{4}\right]+2(y-1)$
$=\frac{\left(y^{2}-8\right) y+8(y-1)}{4}=\frac{y^{3}-8 y+8 y-8}{4}=\frac{y^{3}-8}{4}$
$\Rightarrow f''(y)=\frac{3 y^{2}}{4}$
$f'(y)=0 $
$\Rightarrow \frac{y^{3}-8}{4}=0$
$\Rightarrow y^{3}-8=0$
$\Rightarrow y=2$
$f''(2)=\frac{3.2^{2}}{4}=3>0$.
When $y=2$, the distance between the points is minimum.
$\therefore y=2$
$\Rightarrow x=1 .$
$\therefore $ Required point $=(1,2)$.