If the shortest distance of the point A(0, a) from the parabola $y=x^2$ is $\frac{\sqrt{4c-1}}{2}$, then :

A. $y=\frac{2a-1}{2}$

B. $a=2c$

C. Point A is (0, c)

D. Point A is (0, 2c)

E, $y=\frac{2c-1}{2}$

Choose the correct answer from the options given below :

C, E Only

$\text{Point on parabola } y=x^2 \text{ be }(x,x^2).$

$\text{Distance}^2 = (x-0)^2+(x^2-a)^2.$

$D^2=x^2+(x^2-a)^2.$

$\frac{d}{dx}D^2=2x+2(x^2-a)\cdot2x=0.$

$x\{2(x^2-a)+1\}=0.$

$x^2=a-\frac12.$

$\text{Minimum distance}^2=(a-\frac12)+\frac14=a-\frac14.$

$\text{Given } \text{distance}=\frac{\sqrt{4c-1}}{2}.$

$\Rightarrow \text{distance}^2=\frac{4c-1}{4}.$

$a-\frac14=\frac{4c-1}{4} \Rightarrow a=c.$

$\text{Closest point has }y=a-\frac12=\frac{2c-1}{2}.$

$\text{Correct options are (C) and (E).}$