Standard form of the equation of parabola

Click For Summary
SUMMARY

The standard form of the equation for a parabola with an axis of symmetry along the x-axis (y=0) and a focus at (-5,0) is derived from the definition of a parabola. The correct equation format is x = a(y - k)² + h, where (h, k) is the vertex. Given the focus, the vertex is located at (-5,0), and the equation can be expressed as x = a(y - 0)² - 5. The value of 'a' can be determined based on the distance from the vertex to the focus.

PREREQUISITES
  • Understanding of conic sections, specifically parabolas
  • Familiarity with the vertex form of a parabola equation
  • Knowledge of the focus and directrix properties of parabolas
  • Ability to manipulate quadratic equations
NEXT STEPS
  • Study the vertex form of a parabola equation in detail
  • Learn how to derive the equation of a parabola from its focus and directrix
  • Practice problems involving the conversion of standard form equations of parabolas
  • Explore examples of parabolas with different orientations and their equations
USEFUL FOR

Students learning conic sections, particularly those studying parabolas, as well as educators seeking to provide clear examples of deriving equations from geometric properties.

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.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
9
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K