Find the point on the curve

Solution:

Let point $(x, y)$ be the nearest to the point $(2,-8)$. Then,

$y^{2}=4 x$

$\Rightarrow x=\frac{y^{2}}{4}$                     ......(1)

$d^{2}=(x-2)^{2}+(y+8)^{2}$           [Using distance formula]

Now,

$Z=d^{2}=(x-2)^{2}+(y+8)^{2}$

$\Rightarrow Z=\left(\frac{y^{2}}{4}-2\right)^{2}+(y+8)^{2}$        [From eq. (1)]

$\Rightarrow Z=\frac{y^{4}}{16}+4-y^{2}+y^{2}+64+16 y$         

$\Rightarrow \frac{d Z}{d y}=\frac{4 y^{3}}{16}+16$

For maximum or minimum values of $Z$, we must have

$\frac{d Z}{d y}=0$

$\Rightarrow \frac{4 y^{3}}{16}+16=0$

$\Rightarrow \frac{4 y^{3}}{16}=-16$

$\Rightarrow y^{3}=-64$

$\Rightarrow y=-4$

Substituting the value of $x$ in eq. (1), we get

$x=4$

Now,

$\frac{d^{2} Z}{d y^{2}}=\frac{12 y^{2}}{16}$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=12>0$

So, the nearest point is $(4,-4)$.