Question

Find the shortest distance of the point $(0,c)$ from the parabola $y=x^2$ where $0\le c\le 5$.

Answer 1

Let $x$ be distance between $(0,C)$ and any point$(x,y)$ on the parabola $y=x^2$ ..... (1)

Put value of $y$ in equation (1) and we get

Let

$(x{}_1,y_1)=(x,y)$

$(x{}_2,y_2)=(0,c)$

Using distance formula,

$s=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$s=\sqrt{{(0-x)}^2+(c-y^2)}$

$s=\sqrt{x^2+{(c-y)}^2}$

$s=\sqrt{x^2+{(y-c)}^2}$ ……… (1)

Differentiate this equation w.r.t. $y$, we get,

$\frac{ds}{dy}=\frac{d}{dy}\left(\sqrt{y+{(y-c)}^2}\right)$

Since, $(\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}})$

Therefore,

$\frac{ds}{dy}=\frac{1}{2\sqrt{y+{(y-c)}^2}}\frac{d}{dy}(y+{(y-c)}^2)$

$\frac{ds}{dy}=\frac{1}{2\sqrt{y+{(y-c)}^2}}(1+2(y-c\left)\right)$

$\frac{ds}{dy}=\frac{1+2y-2c}{2\sqrt{y+{(y-c)}^2}}$

For maxima and minima,

$\frac{ds}{dy}=0$

$\frac{1+2y-2c}{2\sqrt{y+{(y-c)}^2}}=0$

$1+2y-2c=0$

$2y=2c-1$

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

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

When, $y<c-\frac{1}{2}$, just $\frac{ds}{dy}=$ negative

And when, $y<c-\frac{1}{2}$, just $\frac{ds}{dy}=$ positive

But at $y=c-\frac{1}{2},\frac{ds}{dy}$ changes from $–ve$ to $+ve$

Hence $s$ is minimum at $y=c-\frac{1}{2}$

By equation $\left(1\right)$, shortest distance

$s=\sqrt{y+{(y-c)}^2}$

$=\sqrt{(c-\frac{1}{2})+{(c-\frac{1}{2}-c)}^2}$

$=\sqrt{(c-\frac{1}{2})+\frac{1}{4}}$

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

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

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

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

Hence, this is the required distance.