Explanation

(a)
Let the required point be $(x,\ y)$ on the curve
So, $d = \sqrt{(x - 4)^{2} + \left( y + \frac{1}{2} \right)^{2}}$ should be minimum
Let $D = (x - 4)^{2} + \left( y + \frac{1}{2} \right)^{2}$
$\therefore\ \ D = (x - 4)^{2} + \left( x^{2} + \frac{1}{2} \right)^{2}$
$\Rightarrow \ \ \ D = x^{4} + 2x^{2} - 8x$
$\therefore\ \ \ D^{'} = 4x^{3} + 4x - 8$
Put $D^{'} = 0$ for maxima or minima
$\therefore\ \ x^{3} + x - 2 = 0\ \ \ \ \Rightarrow \ \ \ \ x = 1$
Now, $D^{''} = 12x^{2} + 4$
at $x = 1,\ \ \ D^{''} = 16 > 0$
$\therefore\ \ D$ is minimum when $x = 1$
Hence, the required point is (1, 1)