Find the point on the curvey

Suppose a point $(\mathrm{x}, \mathrm{y})$ on the curve $y^{2}=2 x$ is nearest to the point $(1,4) .$ Then,

$y^{2}=2 x$

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

$d^{2}=(x-1)^{2}+(y-4)^{2}$           [Using distance formula]

Now,

$Z=d^{2}=(x-1)^{2}+(y-4)^{2}$

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

$\Rightarrow Z=\frac{y^{4}}{4}+1-y^{2}+y^{2}+16-8 y$

$\Rightarrow \frac{d Z}{d y}=y^{3}-8$

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

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

$\Rightarrow y^{3}-8=0$

$\Rightarrow y^{3}=8$

$\Rightarrow y=2$

Substituting the value of $y$ in $(1)$, we get

$x=2$

Now,

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

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

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