IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q16

Let $(t,\ t^2)$ be some point on the parabola.

Then, the distance between the point $(0,\ c)$ and the point $(t,\ t^2)$ parabola is given by $\begin{aligned} d &= \sqrt{(t-0)^2+(t^2-c)^2} \\&= \sqrt{(t^2)^2 + (1-2c)t^2 + c^2} \\&= \sqrt{p^2 + (1-2c)p + c^2} \ ... (p=t^2) \end{aligned}$ which is minimum if (let) $f(p)=p^2 + (1-2c)p + c^2$ is minimum.

The minimum value of the quadratic $f(p)=0$ occurs at $p=\frac{2c-1}{2}$ (if $c>\frac12$ ), when $f'(p)=0$ ; and then, $d=\sqrt{c-\frac14}$ (if $c>\frac14$ ).

$c>\frac12$ and $c>\frac14$ $\implies$ $c>\frac12$ and therefore, $(c,\ d)$ cannot definitely be $\boxed{\text{(C)}\ (\frac5{8}, \frac5{8})}$ and $\boxed{\text{(D)}\ (\frac7{8}, \frac7{8})}$ (and hence, are the answer options).

Further, for $c<\frac12$ , $f'(p)>0 \ \forall \ p=t^2>0$ and therefore, $f(p)$ is an increasing function which attains its minimum when $p$ is minimum, that is $p=t^2=0$ ; and then $d=c$ . Therefore, $(c,\ d)$ can be $\boxed{\text{(A)}\ (\frac1{8}, \frac1{8})}$ and $\boxed{\text{(B)}\ (\frac3{8}, \frac3{8})}$ (and hence are not the answer options).

Thus, the answer is $\boxed{\text{9911}}$ .