The point on the curve

Solution:

$(b)(1,2)$

Let the required point be $(x, y)$. Then,

$y^{2}=4 x$

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

Now,

$d=\sqrt{(x-2)^{2}+(y-1)^{2}}$]

Squaring both sides, we get

$\Rightarrow d^{2}=(x-2)^{2}+(y-1)^{2}$

$\Rightarrow d^{2}=\left(\frac{y^{2}}{4}-2\right)^{2}+(y-1)^{2}$

$\Rightarrow d^{2}=\frac{y^{4}}{16}+4-y^{2}+y^{2}+1-2 y$           [From eq. (1)]

Now,

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

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

$\Rightarrow \frac{d Z}{d y}=\frac{y^{3}}{4}-2$

$\Rightarrow \frac{y^{3}}{4}-2=0$

$\Rightarrow y^{3}=8$

$\Rightarrow y=2$

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

$x=1$

Now,

$\frac{d^{2} Z}{d y^{2}}=\frac{3 y^{2}}{4}$

$\Rightarrow \frac{d^{2} Z}{d y^{2}}=\frac{3(2)^{2}}{4}=3>0$

So, the nearest point is $(1,2)$.