Find the shortest distance of the point $C(0, c)$ from the parabola $y = x^2$.

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

The correct answer is Option (1) → $\frac{\sqrt{4c-1}}{2}$

Let $P(x, y)$ be any point on the parabola $y = x^2$, then

$|CP| = \sqrt{(x –0)^2 + (y - c)^2} = \sqrt{y + (y - c)^2}$   (using $x^2=y$)

$=\sqrt{y^2 - (2 c − 1) y + c^2}$.

Now $|CP|$ is shortest iff $|CP|^2$ is shortest.

Let us write $|CP|^2$ as $f(y)$ i.e.

$f(y) = y^2 - (2c - 1) y + c^2$   ...(i)

Differentiating (i) w.r.t. y, we get

$f'(y) = 2y - (2c - 1)$ and $f''(y) = 2$.

$f'(y) = 0 ⇒ 2y - (2c - 1) = 0$

$⇒y=\frac{2c-1}{2}$

For $y=\frac{2c-1}{2},f''(y)=2>0$

$⇒f(y)$ is minimum when $y=\frac{2c-1}{2}$ i.e. $|CP|$ is minimum when $y=\frac{2c-1}{2}$ and the minimum value of $|CP|=\sqrt{\frac{2c-1}{2}+\left(\frac{2c-1}{2}-c\right)^2}=\sqrt{\frac{2c-1}{2}+\frac{1}{4}}=\frac{\sqrt{4c-1}}{2}$.