# How do you find the equation of a parabola when given two points (2,10) and (4,10)?

Jul 26, 2015

There's not enough information here to uniquely determine the parabola, but enough to find:

$y = a {x}^{2} - 6 a x + \left(8 a + 10\right)$ for some constant $a \in \mathbb{R}$

#### Explanation:

Since the $y$ coordinate of both points is the same, the axis must be midway between the two points at $x = 3$.

So the equation of the parabola takes the form:

$y = a {\left(x - 3\right)}^{2} + k$

If we substitute $x = 4$ and $y = 10$ into this equation we get:

$10 = a {\left(4 - 3\right)}^{2} + k = a + k$

So $k = 10 - a$ and we can write the equation as:

$y = a {\left(x - 3\right)}^{2} + \left(10 - a\right)$

$= a {x}^{2} - 6 a x + 9 a + 10 - a$

$= a {x}^{2} - 6 a x + \left(8 a + 10\right)$

The constant $a$ can be given any value in $\mathbb{R}$, except that when $a = 0$ the 'parabola' is the straight line $y = 10$.

Here are the parabolas for $a = \pm 1$ and $a = \pm 2$

graph{(y-x^2+6x-18)(y+x^2-6x-2)(y-2x^2+12x-26)(y+2x^2-12x+6)=0 [-2.585, 7.415, 7.22, 12.22]}