What Is the Equation of a Parabola with Vertex (-2, 3) and Focus (0, 3)?

  • Thread starter Dirac
  • Start date
  • Tags
    Focus
In summary: I already said you a is the distance between the focus and vertex.here you goa=sqrt[(0-(-2))^2+(3-3)]=sqrt[2^2]
  • #1
Dirac
19
0
vertex (-2, 3), focus (0,3)

Find the equation of the parabola.

Dirac.
 
Physics news on Phys.org
  • #2
Dirac said:
vertex (-2, 3), focus (0,3)

Find the equation of the parabola.

Dirac.
What have you tried so far?
Find the directix. then recall
A parabola is the set of point equidistant from the focus and directix.
 
Last edited:
  • #3
lurflurf said:
What have you tried so far?
Find the directix. then recall
A parabola is the set of point equidistant from the focus and directix.

So far I have got to this

((y-3)^2)=4a(x+2)

What now

Dirac.
 
  • #4
Dirac said:
So far I have got to this

((y-3)^2)=4a(x+2)

What now

Dirac.
a will be the distance between the focus and vertex.
Why is this?
 
  • #5
lurflurf said:
a will be the distance between the focus and vertex.
Why is this?

How does one find the vertex
 
  • #6
Dirac said:
How does one find the vertex
you started out with
vertex (-2, 3), focus (0,3)
The points of a parabola are equidistant from it directrix and focus.
The vertex is the point closest to them. that is
distance(vertex,focus)<=distance(point on parabola,focus)
with equality only when the point is the vertex.
 
  • #7
lurflurf said:
you started out with
vertex (-2, 3), focus (0,3)
The points of a parabola are equidistant from it directrix and focus.
The vertex is the point closest to them. that is
distance(vertex,focus)<=distance(point on parabola,focus)
with equality only when the point is the vertex.

Could you just post the answer, or would it get up your backside to post something useful in your life

Dirac.
 
  • #8
Dirac said:
Could you just post the answer, or would it get up your backside to post something useful in your life

Dirac.
I already said you a is the distance between
vertex (-2, 3), focus (0,3)
here you go
a=sqrt[(0-(-2))^2+(3-3)]
=sqrt[2^2]
a=2
Do you understand why?
Can you find the equation of a parabola with
vertex (0,0), focus (0,-4)
 
  • #9
Dirac said:
Could you just post the answer, or would it get up your backside to post something useful in your life

Dirac.

How pathetic. Why in the world would anyone care what the answer to some made up question like this is? The whole point of an exercise like this is to learn how to get the answer. I'll bet there's a "general formula" for the parabola in the same section of your text as this problem. Look it up. We're here for people who want to learn not for people who want the answers handed to them.
 

Related to What Is the Equation of a Parabola with Vertex (-2, 3) and Focus (0, 3)?

What is a directrix and focus in mathematics?

A directrix and focus are two important elements in conic sections, such as circles, ellipses, parabolas, and hyperbolas. They are used to define the shape and position of these curves in a coordinate plane.

How are the directrix and focus related in conic sections?

In conic sections, the directrix is a fixed line that is perpendicular to the axis of symmetry of the curve. The focus is a fixed point on the axis of symmetry. The distance from the focus to any point on the curve is equal to the distance from that point to the directrix. This relationship helps determine the shape of the conic section.

What is the role of the directrix and focus in a parabola?

In a parabola, the directrix is a line that is parallel to the axis of symmetry and is located at a distance from the vertex equal to the distance from the vertex to the focus. The focus is the fixed point on the axis of symmetry that gives the parabola its shape. The directrix and focus are used in the standard equation of a parabola: y = (1/4p)(x-h)^2 + k, where p is the distance from the focus to the vertex.

How do the directrix and focus affect the eccentricity of an ellipse?

The eccentricity of an ellipse is a measure of how "stretched out" or "squished" the ellipse is. It is determined by the distance between the center of the ellipse and the two foci, divided by the length of the major axis. The closer together the foci are, the higher the eccentricity. The directrix and focus play a key role in determining the eccentricity of an ellipse.

Can the directrix and focus be used to find the equation of a hyperbola?

Yes, the directrix and focus can be used to find the equation of a hyperbola. In a hyperbola, the distance from any point on the curve to the focus is always greater than the distance from that point to the directrix. This relationship helps determine the shape and position of the hyperbola in a coordinate plane. The standard equation of a hyperbola is (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where a is the distance from the center to a vertex, and b is the distance from the center to a co-vertex, both measured along the transverse axis.

Similar threads

Replies
44
Views
3K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
292
  • Calculus and Beyond Homework Help
Replies
21
Views
1K
Replies
7
Views
3K
Replies
4
Views
814
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
Back
Top