Standard form of the equation of parabola

AI Thread Summary
To write the equation of a parabola with an axis of symmetry along y=0 and a focus at (-5,0), the standard form is x = ay^2 + by + c, as the parabola opens horizontally. The vertex lies on the axis of symmetry, and the focus and vertex are aligned along this axis. The equation x = -b/(2a) helps find the vertex's x-coordinate, but it is not the equation of the parabola itself. The discussion emphasizes the need for a worked example to clarify the application of these concepts. Understanding conics requires practice, and finding similar examples in textbooks can aid in grasping the material.
jOANIE
Messages
6
Reaction score
0

Homework Statement



Write the equation in standard form for the parabola with an axis of symmetry y=0 and focus(-5,0).

Homework Equations


I think ax^2+bx+c. Also maybe x = -b/2a. But I don't know how to apply these.



The Attempt at a Solution


I know that focus and vertex lie on axis of symmetry and that directrix ane axis of symmetry are perpendicular to each other.

If you could give me a similar example worked out, I would be able to do this one. Thanks.
 
Physics news on Phys.org
jOANIE said:

Homework Statement



Write the equation in standard form for the parabola with an axis of symmetry y=0 and focus(-5,0).

Homework Equations


I think ax^2+bx+c.
I think not. First off, that's not an equation. Second, the axis of symmetry is the line y = 0 (the x-axis). This means that the equation will be x = ay^2 + by + c.
jOANIE said:
Also maybe x = -b/2a. But I don't know how to apply these.
x = -b/(2a) isn't the equation of a parabola; it gives you the x-coordinate of the vertex for a parabola that opens up or down. Your parabola opens left or right.
jOANIE said:

The Attempt at a Solution


I know that focus and vertex lie on axis of symmetry and that directrix ane axis of symmetry are perpendicular to each other.

If you could give me a similar example worked out, I would be able to do this one. Thanks.

Based on the information you provided, there are an infinite number of parabolas that satisfy these conditions. Aren't there any examples showing how to do this in your textbook?
 
Hello--Thanks for responding. I cannot find an example like this one in my book. Could you give me any hints on how to begin this problem? I am completely self-taught in math and am trying to understand conics, which I have not attempted before. I am also good at following examples and, actually, learn quite a bit from them.
 
OK, what's the example that seems the closest? Are there any examples where you're supposed to find the equation of a parabola of any kind? Maybe we can help you make some connections.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top