Two Parabolas with Vertices on the x-axis

[turophile]

Here's the statement of the problem:

A parabola with vertical axis has its vertex on the x-axis and passes through the points (2, 3) and (3, 12). What is its equation?

I made a sketch, and there are two parabolas that fit this description. One parabola includes the two (non-vertex) points on one of its "arms", and the same points are on separate arms of the other parabola.

Here's what I've calculated so far:

Let A be the set of points on the parabola, and let the vertex of the parabola be (x_v, 0).
(2, 3) ? A ? 3 – k = a(2 – h)^2 = a(4 – 4h + h^2) = 4a – 4ah + ah^2 ? k = 3 – 4a + 4ah – ah^2 [1]
(3, 12) ? A ? 12 – k = a(3 – h)^2 = a(9 – 6h + h^2) = 9a – ah + ah^2 ? k = 12 – 9a + ah – ah^2 [2]
(x_v, 0) ? A ? 0 – k = a(x_v – h)^2 = a((x_v)^2 – 2(x_v)h + h^2) = a(x_v)^2 – 2a(x_v)h + ah^2
? k = – a(x_v)^2 + 2a(x_v)h – ah^2 [3]
[1] and [2] ? 3 – 4a + 4ah – ah^2 = 12 – 9a + ah – ah^2 ? 9a – 4a + 4ah – ah = 12 – 3 ? 5a + 3ah = 9 [4]
[1] and [3] ? 3 – 4a + 4ah – ah^2 = – a(x_v)^2 + 2a(x_v)h – ah^2 ?

At this point, I'm getting a little lost. Am I on the right track? Any hints for proceeding? Thanks.

 

[soroban]

Hello, turophile!

A parabola with vertical axis has its vertex on the x-axis
and passes through the points (2, 3) and (3, 12).
What is its equation?

\(\displaystyle \text{A "vertical" parabola has the equation: }\;(x-h)^2 \:=\:a(y-k)\)
. . \(\displaystyle \text{where }(h,k)\text{ is the vertex.}\)

\(\displaystyle \text{Our parabola has vertex }(h,0)\)
\(\displaystyle \text{Its equation is: }\;(x-h)^2 \:=\:ay\)


\(\displaystyle \text{Substitute the given points:}\)

. . \(\displaystyle \begin{array}{cccccc}(2,3)\!: & (2-h)^2 &=& 3a & [1] \\ (3,12)\!: & (3-h)^2 &=& 12a & [2] \end{array}\)


\(\displaystyle \text{Divide [1] by [2]: }\;\frac{(2-h)^2}{(3-h)^2} \:=\:\frac{3a}{12a} \quad\Rightarrow\quad \frac{4-4h+h^2}{9-6h-h^2} \:=\:\frac{1}{4}\)

. . . \(\displaystyle 16 - 16h + 4h^2 \:=\:9 - 6h + h^2 \quad\Rightarrow\quad 3h^2 - 10h + 7 \:=\:0\)

. . . . . . . . \(\displaystyle (h-1)(3h-7) \:=\:0 \quad\Rightarrow\quad h \:=\:1,\:\tfrac{7}{3}\)


Substitute into [1]:

. . \(\displaystyle \begin{array}{ccccccccc}h=1\!: & (2-1)^2 &=& 3a & \Rightarrow & a &=& \frac{1}{3} \\ \\[-3mm] h = \frac{7}{3}: & (2-\frac{7}{3})^2 &=& 3a & \Rightarrow & a &=& \frac{1}{27} \end{array}\)


\(\displaystyle \text{There are two parabolas: }\;\begin{Bmatrix}(x - 1)^2 &=& \frac{1}{3}y \\ \\[-3mm] \left(x-\frac{7}{3}\right)^2 &=& \frac{1}{27}y \end{Bmatrix}\)