Determine the points on the curve

Solution:

Let the point $(x, y)$ on the curve $x^{2}=4 y$ be nearest to $(0,5)$. Then,

$x^{2}=4 y$

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

Also,

$d^{2}=(x)^{2}+(y-5)^{2}$        [Using distance formula]

Now,

$Z=d^{2}=(x)^{2}+(y-5)^{2}$

$\Rightarrow Z=(x)^{2}+\left(\frac{x^{2}}{4}-5\right)^{2}$      $\left[\begin{array}{ll}\text { Using eq. } & \text { (1) }\end{array}\right]$

$\Rightarrow Z=x^{2}+\frac{x^{4}}{16}+25-\frac{5 x^{2}}{2}$

$\Rightarrow \frac{d Z}{d y}=2 x+\frac{4 x^{3}}{16}-5 x$

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

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

$\Rightarrow 2 x+\frac{4 x^{3}}{16}-5 x=0$

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

$\Rightarrow x^{3}=12 x$

$\Rightarrow x^{2}=12$

$\Rightarrow x=\pm 2 \sqrt{3}$

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

$y=3$

Now,

$\frac{d^{2} Z}{d y^{2}}=2+\frac{12 x^{2}}{16}-5$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=9-3=6>0$

So, the required nearest point is $(\pm 2 \sqrt{3}, 3)$.