# Find the equation of the parabola that contains points (1,10) & (2,4), has a vertical axis of symmetry, and whose vertex is on the line $4x-3y=6$.

Find the equation of the parabola which contains the points $(1,10)$ and $(2,4)$, has a vertical axis of symmetry and whose vertex is on the line $4x-3y=6$.

What approach should be taken here?

you have two equations $$a+b+c=10$$ and $$4a+2b+c=4$$ and since the vertex is on the line $$y=\frac{4}{3}x-2$$ the equation $$\frac{4ac-b^2}{4ac}=\frac{4}{3}\cdot \left(-\frac{b}{2a}\right)-2$$

You are right, there doesn't seem to be enough information. There actually exist two parabolas that work:

$$y = 2(x-3)^2 + 2$$ $$y = 26(x - \frac{21}{13})^2 + \frac{2}{13}$$

You can get these results by using the vertex formula for the parabola

$$y = a(x-d)^2 + e$$

where $(d,e)$ is the vertex and $a$ is a stretching factor. Thus, the provided information gives us the following

$$a(d-1)^2 + e = 10$$ $$a(d-2)^2 + e = 4$$ $$4d - 3e = 6$$

Full solution:

Subtracting the first two equations yields
$$a = \frac{6}{2d-3}$$
and the third equation yields
$$e = \frac{4}{3}d - 2$$
So substitute $a$ and $e$ in
$$a(d-1)^2 + e = 10$$
to get
$$\frac{6}{2d-3} (d-1)^2 + \frac{4}{3}d - 2 = 10 .$$
After simplifying, this becomes a quadratic
$$\frac{26}{3}d^2 - 40d + 42 = 0$$
which has solutions $d = \frac{21}{3}$ and $d = 3$. Plug these back in to get the respective values for $e$ and $a$. As it turns out, both produce valid parabolas.
Thus, there are two such parabolas. One with vertex $(3,2)$ and one with vertex $(\frac{21}{13}, \frac{2}{13})$.