Squaring a function, getting a surprising result

Question

I'm doing a problem in which I have to find the minimum distance between the origin and a point on a parabola. I understand that the minimum of a function and the minimum of the square of the function are the same. The distance function when considering that one point is the origin should be $d =\sqrt{x^2 + y^2}$, making the squared function $d = x^2 + y^2$. However, when solving for $y$ so that I can make the equation solely in terms of $x$, I get $y = \sqrt{-x^2}$, and substituting that for $y$ in the original equation would negate that square root and result in $x^2 -x^2$, or $0$. Hopefully I just overlooked something, but that obviously seems wrong if I'm trying to find a distance function I can derive and get a minimum.

Answer 1

The general point on the parabola $y=x^2$ has two coordinates: $(t,t^2)$ for $t$ a real. Use this in the distance squared to $(0,0)$ formula. Multiply out, find derivative, ETC.

I don't think parabola formula is $y=x^2$ since then min distance clearly zero (no work to do). But same idea works if $y=ax^2+bx+c$ where $a \neq 0.$ then general point is $(t,at^2+bt+c)$ and proceed as above.

Note: It just struck me that your problem may have been in using $x,y$ for both the point on the parabola, and in applying $x^2+y^2$ for distance to origin. If you keep $x,y$ for the parabola point, then must solve for one of $x,y$ in terms of the other to set things up. That's why a good idea is to use another variable like $t$ to get the parabola parametrized in that new letter, like $(t,t^2+1)$ or what have you. Initially check if $(0,0)$ already on the parabola [if so stop, min distance $0.$] Otherwise in $d^2=x^2+y^2$ you have to substitute the coordinates into that, e.g. in my made-up example $t^2+(t^2+1)^2.$ Good luck.

I looked at your example $x^2-2x+5$ and found the same numeric value for $x$ as you mentioned in comments. Then I applied the cubic formula and got that, if $u=\frac{-1+\sqrt{55}}{4},$ then $$x=u^{1/3}+\frac{3}{2u^{1/3}}+1.$$ A google search should get you to several versions of the cubic formula.

Answer 2

If you want to find the distance between a point on the parabola and the origin, all you have to do is use the Pythagorean theorem (essentially the distance equation you used) to figure out the hypotenuse connecting the points. For example, if the point on the parabola is $(3,9)$, then the distance between $(0,0)$ and $(3,9)$ is $\sqrt{3^2+9^2}=\sqrt{90}=3\sqrt{10}$