# Prove that three points define a unique parabola

How do we prove that there is always a unique parabola (with equation $y=ax^2+bx+c$) that passes through 3 distinct points $P_1 (p_1,q_1), P_2 (p_2,q_2), P_3(p_3,q_3)$ ?

If I choose to use matrices and row reduction, how do I deal with the fact that the coordinates might be zero?

• First prove that there is a parabola that passes through $(p_1,q_1)$, $(p_2,0)$, and $(p_3,0)$. (Once you have that, you also know there's one through $(p_1,0)$, $(p_2,q_2)$, and $(p_3,0)$ by relabeling, and similarly for the last coordinate; then you can just add the equations together.) – Akiva Weinberger Aug 31 '15 at 1:57
• The statement is only true if you assume your three points are not colinear. Otherwise, consider the points $(0,0),(-1,0),(1,0)$. Clearly there is no such parabola as that would be a quadratic equation with 3 roots. (Unless you consider a straight line to be a degenerate parabola, which generally it isn't :)) – Alan Aug 31 '15 at 1:58
• @Alan Technically, that's $y=0x^2+0x+0$, though I see your point. – Akiva Weinberger Aug 31 '15 at 1:59
• OP, you should prove that there is no (proper) parabola through $(-1,0)$, $(0,0)$, and $(1,0)$, by the way. – Akiva Weinberger Aug 31 '15 at 2:00
• Not is unique frodriguezdiaz.blogspot.mx/2013/11/… – jimbo Aug 31 '15 at 2:13

Proving that there is a parabola should be no hassle. To show uniqueness, suppose $y = a_1x^2 + b_1x + c_1$ and $y = a_2x^2 + b_2x + c_2$ fit the points. Then the parabola $y = (a_2 - a_1)x^2 + (b_2 - b_1)x + (c_2 - c_1)$ goes through the points $(p_1,0),(p_2,0),(p_3,0)$. This implies the quadratic has three roots, which is only possible if $a_1 = a_2$, $b_1 = b_2$, and $c_1 = c_2$.
You have a set of points $(p_i,q_i)$ for $i=1,2,3$ with $p_i\neq p_j$ for $i\neq j$.
You can prove existence of a solution by deriving the determinant of a Vandermonde matrix of size 3: $\prod_{1\leq i<j\leq 3} (p_i-p_j)$ and seeing that it is not singular from your assumptions above.