Question

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

• A. $c$ if $0\le c\le \frac{1}{2}$
• B. $c$ if $3\le c\le 5$
• C. $\sqrt{c-\frac{1}{4}}$ if $\frac{1}{2}\le c\le 5,c$ if $0\le c\le \frac{1}{2}$
• D. $\sqrt{c}$

Answer 1

The correct option is C $\sqrt{c-\frac{1}{4}}$ if $\frac{1}{2}\le c\le 5,c$ if $0\le c\le \frac{1}{2}$

It S is the distance of the point (x, y) on the parabola $y=x^2$ from (0, c). Then

$S=\sqrt{x^2+{(y-c)}^2}=\sqrt{x^2+{(x^2-c)}^2}$ (1)

which is least $\Rightarrow S^2$ is least

$\frac{d{S}^2}{dx}=2x+2(x^2-c)\times 2x=0$

$\Rightarrow x=0or{x}^2=c-1/2$

when$x=0,S=c,$ when $x^2=c-1/2$, $c\ge 1/2$

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

Hence the least distance is

c if $0\le c\le 1/2$ and $\sqrt{c-\frac{1}{4}}$ if $\frac{1}{2}\le c\le 5$