TITLE: 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 $4x-3y=6$.
QUESTION [0 upvotes]: 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 $4x-3y=6$.
I'm not sure how to go about this. I feel there is not enough information. 
What approach should be taken here?

REPLY [1 votes]: 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 $(\frac{21}{13}, \frac{2}{13})$.
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