# Thread: How do I find the equation of the parabola given the information in the diagram?

1. ## How do I find the equation of the parabola given the information in the diagram?

With the information in the diagram, how do I find the equation of the parabola? Is it even possible or do I need more information?

2. ## Re: How do I find the equation of the parabola given the information in the diagram?

Your diagram leaves much to be desired ... I interpret it to mean a parabola that has x-intercepts at (0,0) and (0,10) where the slope of the tangent to the parabola makes a 30 degree angle with the x-axis at (0,0).

If that is the case, then see the attached graph. If you mean something else entirely (in other words, my interpretation is completely wrong), then you’ll have to provide a more concise explanation of what you are looking for ...

3. ## Re: How do I find the equation of the parabola given the information in the diagram?

I'm glad you were able to get that information from the picture, skeeter. I thought it was (0, N) rather than (0, 10) and completely missed the angle the first time I looked!

The fact this parabola has horizontal line of symmetry and goes through (0, 0) and (0, 10) means it can be written $x= a(y-0)(y- 10)= ay^2- 10ay$. The derivative at (0, 0) is $x'= 2a(0)- 10a= -10a= tan(30)$.

4. ## Re: How do I find the equation of the parabola given the information in the diagram?

my ipad turned the pic sideways making me think the axis was horizontal ... looks like y-intercepts (0,0) and (0,10) instead of what I interpreted

revised graph attached

5. ## Re: How do I find the equation of the parabola given the information in the diagram?

Originally Posted by HallsofIvy
I'm glad you were able to get that information from the picture, skeeter. I thought it was (0, N) rather than (0, 10) and completely missed the angle the first time I looked!

The fact this parabola has horizontal line of symmetry and goes through (0, 0) and (0, 10) means it can be written $x= a(y-0)(y- 10)= ay^2- 10ay$. The derivative at (0, 0) is $x'= 2a(0)- 10a= -10a= tan(30)$.
You may be reversing the x-axis. I believe you want $x' = 2a(0)-10a = -10a = -\tan(30^\circ) \Longrightarrow a = \dfrac{\sqrt{3}}{30}$

So, the function would be:

$x = \dfrac{\sqrt{3}}{30}y^2-\dfrac{\sqrt{3}}{3}y$

6. ## Re: How do I find the equation of the parabola given the information in the diagram?

Originally Posted by SlipEternal

So, the > > function < < would be:

$x = \dfrac{\sqrt{3}}{30}y^2-\dfrac{\sqrt{3}}{3}y$

That would be a relation, but, except at the vertex, there are two y-values for every x-value.

7. ## Re: How do I find the equation of the parabola given the information in the diagram?

Originally Posted by greg1313
That would be a relation, but, except at the vertex, there are two y-values for every x-value.
$x = f(y)$