Shortest distance of the point $(0,c)$ from the parabola $y=x^2$ ? (Where $0\le c \le5$)

My approach: I wrote the distance formula by taking parametric coordinates as $(t,t^2)$ and then differentiated the equation.I got the extremum as $x=\sqrt (c)$,but that's the wrong answer? I can't figure out my mistake.

Please help!

  • $\begingroup$ i think maximum is coming at t=0 $\endgroup$
    – Taylor Ted
    Jul 10, 2015 at 6:16

4 Answers 4


Let $f(t):=(t-0)^2+(t^2-c)^2$. The necessary condition for the minimum of $f$ is $$ \frac{d}{dt}\left\{(t-0)^2+(t^2-c)^2\right\}=2t+4t(t^2-c)=0, $$

which yields

$$ t_1=0, t_{2,3}=\pm\sqrt{c-1/2}. $$

However, $t_{2,3}$ are real only for $c\ge1/2$. In addition, the second order condition shows that when $c\ge 1/2$, $f(t_1)$ is a local maximum. So, the minimum distance squared is $c^2$ when $c<1/2$ and $c-1/4$, otherwise.


I read somewhere that the vector from the point to the curve must be normal to the point of the curve which it is closest to. I think you should read "differential geometry" if you think this is exciting.

We can get tangents to $x^2$ by differentiation: it is the line passing through $(x,x^2)$ having slope $2x$. Then the normal will pass through the same point but have inverted slope: $-1/(2x)$. We can convince ourselves by calculating and see that $[1,2x] [1,-1/(2x)]^T = 0$, so they are orthogonal.

So rays shot out from the curves' normal needs to hit the point. We can reformulate this as rays with slope $-1/(2x_0)$ shot out from the point need to hit $(x_0,{x_0}^2)$. So

$$c - 1/(2x_0) x_0 = {x_0}^2 \Leftrightarrow$$ $$c - 1/2 = {x_0}^2$$ In the plot below, c is on the x-axis and y-value is distance. Blue is the minimum found by brute force in Octave and the red circles are the distances predicted by our formula. enter image description here


The distance from $(0,c)$ to a point on $y=x^2$ is given by the distance formula $$\sqrt{(x-0)^2+(y-c)^2}=\sqrt{x^2+(x^2-c)^2}=\sqrt{x^4+(1-2c)x^2}$$

If we are interested in the shortest distance, it suffices to find the shortest distance of the square of this expression.

Let $$f(x)=x^4+(1-2c)x^2$$ $$f'(x)=4x^3+(2-4c)x$$ and $f'(x)=0$ when $4x^2=2-4c$ or $x=±\sqrt{\frac{1}{2}-c}$

One of these values will give the minimum of the distance; you can probably take it form here.

EDIT: mistakes shown below in the comments; please note.

  • $\begingroup$ There is a mistake in the second last step! $\endgroup$
    – user220382
    Jul 10, 2015 at 6:32
  • $\begingroup$ You are dropping a $c^2$ when rearranging inside the sqrt. $\endgroup$ Jul 10, 2015 at 7:47
  • $\begingroup$ Indeed you forgot a $c^2$ in the last square root. That doesn't alter the result since you then differentiate and that constant vanishes. The argument of minimizing the radicand works fine, though you got the $f'(x)=0$ equality incorrectly rewritten. It should be $4x^2+2-4c=0$, implying $4x^2=4c-2$ and $x=\pm\sqrt{c-\frac12}$. Also, you are dropping the $x=0$ solution. $\endgroup$
    – MickG
    Jul 10, 2015 at 8:14

Given is

$$ y = x^2 $$

So the normal for point $(x_o,x_o^2)$ can be written as

$$ (x,y) = (x_o - 2\alpha x_o, x_o^2 + \alpha ) $$

The length of the normal from point $(x_o,x_o^2)$ is given by

$$ \ell = \alpha \sqrt{4 x_o^2 + 1} $$

We need to solve

$$ (0,c) = (x_o - 2\alpha x_o, x_o^2 + \alpha ) $$


$$ \left[ \begin{array}{rcl} x_o - 2 \alpha x_o &=& 0\\ c &=& x_o^2 + \alpha \end{array} \right. $$


$$ x_o - 2 \alpha x_o = 0 $$

follows $x_o$ or $\alpha = \tfrac{1}{2}$.

Case $x_o=0$

We obtain

$$ \left[ \begin{array}{rcl} 0 &=& 0\\ c &=& \alpha \end{array} \right. $$

Thus $\alpha = c$ and $x_o=0$. So the length is given by

$$ \ell = c $$

Case $\alpha = \tfrac{1}{2}$

We obtain

$$ \left[ \begin{array}{rcl} 0 &=& 0\\ c &=& x_o^2 + \frac{1}{2} \end{array} \right. $$

Thus $\alpha = \tfrac{1}{2}$ and $x_o = \sqrt{c - \tfrac{1}{2}}$. The case $c<\tfrac{1}{2}$ givens the $x_o=0$ solution.

The case $c \ge \frac{1}{2}$ gives the length

$$ \ell = \sqrt { c^2 - \tfrac{1}{4} } $$

Note that $c \ge \sqrt{ c^2 - \tfrac{1}{4}}$ for $c \ge \frac{1}{2}$, so the shortest distance is given by

$$ c \textrm{ for $c \le \tfrac{1}{2}$} $$


$$ \sqrt{c^2 - \tfrac{1}{4}} \textrm{ for $c \ge \tfrac{1}{2}$} $$


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