conic sections
I can't seem to figure this problem out, there doesn't seem to be enough information.
Find the standard equation of the parabola that has a vertical axis
that has $x$-intercepts $-5$ and $3$ and has a lowest point with a $y$-coordinate of $-7$.
I could either do $0 = a(-5 -h)^2 -7$ or $0 = a(3 – h)^2 – 7$ but either way it seems like I don't have enough information. I guess I could just find $a$ in terms of $h$ or viceversa but I feel like there is an easier solution. Any advice?
PERHAPS THIS HELPS
I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you.
You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the vertical axis and a Dandelin sphere. Tangent to this sphere there is a blue plane whose edge can be seen in the figure:
The blue plane and the black cone intersect in a parabola. Also, the blue plane intersects with the other cone in a hyperbola. The parabola's focus and one of the foci of the hyperbola are common.
My conjecture is that this is how to imagine your couple of parabola and hyperbola with a common focus.
Then, I don't know how to proceed.
You have to have more information than that. You can construct infinite number of parabolas with arbitrary x-axis intersects (zero, one or two).
For example you can write $y = (x-x_0)(x-x_1)$ toget a parabola that intersects at $x_1$ and $x_2$, but there are more ways than than. You could for example take $x = y^2+3+\sqrt7$ which is also a parabola - just that it doesn't happen to be facing upwards or downwards.
If you on the other hand requires the parabola to be expressable as $y = P(x)$ where $P$ is an polynomial with rational coefficioents then the solution is unique. The polinomial is uniquely determined up to rational scaling.
Suppose that the coefficient of the $x^2$ term is one, then we have
$$P(x) = (x+3+\sqrt7)(x-c) = x^2 + (3 + \sqrt7-c)x+(3+\sqrt7)cx$$
Now we require $3+\sqrt7-c$ to be rational which makes $c = q+\sqrt7$ for some rational $q$, but we also require $(3+\sqrt7)c = (3+\sqrt7)(q+\sqrt7) = 3q + (3+q)\sqrt7 + 49$ to be rational. But since $\sqrt7$ is irrational we must have that $3+q=0$ and consequently that $c = q+\sqrt7 = -3+\sqrt 7$.
Best Answer
Use the general equation $y=a(x-h)^2+k$, where $(h,k)$ are the coordinates of the vertex.
You already know that the $y$-coordinate of the vertex is $k=-7$. The $x$-coordinate of the vertex, $h$, is given to you indirectly: it is by symmetry the midpoint of the $x$-intercepts. So, find the value of $h$ using this and then substitute the known values of $h$ and $k$ into the general equation.
This still leaves you the unknown $a$. But, you know that the point $(3,0)$ is on the parabola, and you can substitute $x=3$ and $y=0$ into your equation and solve for $a$.