Find the point on the curve

Solution:

Let $(x, y)$ be nearest to the point $(2,4)$. Then,

$x^{2}=8 y$

$\Rightarrow y=\frac{x^{2}}{8}$              .....(1)

$d^{2}=(x-2)^{2}+(y-4)^{2}$        [Using distance formula]

Now,

$Z=d^{2}=(x-2)^{2}+(y-4)^{2}$

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

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

$\Rightarrow \frac{d Z}{d y}=-4+\frac{4 x^{3}}{64}$

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

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

$\Rightarrow-4+\frac{4 x^{3}}{64}=0$

$\Rightarrow \frac{x^{3}}{16}=4$

$\Rightarrow x^{3}=64$

$\Rightarrow x=4$

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

$y=2$

Now,

$\frac{d^{2} Z}{d y^{2}}=\frac{12 x^{2}}{64}$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=3>0$

So, the nearest point is $(4,2)$.