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 :

A, C, E Only

The correct answer is Option 3: A, C, E Only

Let a point on the parabola be $(x, y)$. Since $y = x^2$, the point is $(x, x^2)$.

The square of the distance ($d^2$) from $A(0, a)$ is: $d^2 = (x - 0)^2 + (x^2 - a)^2$ i.e.  $d^2 = x^2 + (x^2 - a)^2$

2. Minimize the Distance 

Let $u = x^2$.  

$d^2 = u + (u - a)^2$ 

$d^2 = u + u^2 - 2au + a^2$ 

$d^2 = u^2 + (1 - 2a)u + a^2$ 

To find the minimum, we differentiate with respect to $u$ and set it to zero: 

$\frac{d(d^2)}{du} = 2u + 1 - 2a = 0$ 

$2u = 2a - 1$ 

$u = \frac{2a - 1}{2}$ 

Since $u = x^2 = y$, we get: 

$y = \frac{2a - 1}{2}$ (This is Statement A)

3. Calculate the Shortest Distance

Substitute $u = \frac{2a - 1}{2}$ back into the $d^2$ equation:

$d^2 = (\frac{2a - 1}{2})^2 + (1 - 2a)(\frac{2a - 1}{2}) + a^2$

$d = \frac{\sqrt{4a - 1}}{2}$

4. Compare with the Given Value 

The question states the shortest distance is $\frac{\sqrt{4c - 1}}{2}$. Comparing the two expressions: 

$\frac{\sqrt{4a - 1}}{2}$ = $\frac{\sqrt{4c - 1}}{2}$

                                     = $a = c$,

• Since $a = c$, the point $A(0, a)$ is actually $(0, c)$ (Statement C).

• Substituting $a = c$ into our expression for $y$: $y = \frac{2c - 1}{2}$ (Statement E).