Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra


powered by FreeFind

 

Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Conics: Ellipses:
    Finding the Equation from Information
(page 3 of 4)

Sections: Introduction, Finding information from the equation, Finding the equation from information, Word Problems


You'll also need to work the other way, finding the equation for an ellipse from a list of its properties.

  • Write an equation for the ellipse having one focus at (0, 3), a vertex at (0, 4),
    and its center at
    (0, 0).
  • Since the focus and vertex are above and below each other, rather than side by side, I know that this ellipse must be taller than it is wide. Then a2 will go with the y part of the equation. Also, since the focus is 3 units above the center, then c = 3; since the vertex is 4 units above, then a = 4. The equation b2 = a2 c2 gives me 16 9 = 7 = b2. (Since I wasn't asked for the length of the minor axis or the location of the co-vertices, I don't need the value of b itself.) Then my equation is:

      y^2 / 16 + x^2 / 7 = 1

  • Write an equation for the ellipse with vertices (4, 0) and (2, 0)
    and foci
    (3, 0) and (1, 0). Copyright Elizabeth Stapel 2010-2011 All Rights Reserved
  • The center is midway between the two foci, so (h, k) = (1, 0), by the Midpoint Formula. Each focus is 2 units from the center, so c = 2. The vertices are 3 units from the center, so a = 3. Also, the foci and vertices are to the left and right of each other, so this ellipse is wider than it is tall, and a2 will go with the x part of the ellipse equation.

    The equation b2 = a2 c2 gives me 9 4 = 5 = b2, and this is all I need to create my equation:

      ((x - 1)^2) / 9 + y^2 / 5 = 1

  • Write an equation for the ellipse centered at the origin, having a vertex at (0, 5) and containing the point (2, 4).
  • Since the vertex is 5 units below the center, then this vertex is taller than it is wide, and the a2 will go with the y part of the equation. Also, a = 5, so a2 = 25. I know that b2 = a2 c2, but I don't know the values of b or c. However, I do have the values of h, k, and a, and also a set of values for x and y, those values being the point they gave me on the ellipse. So I'll set up the equation with everything I've got so far, and solve for b.

     

    ADVERTISEMENT

     

      (y^2)/25 + (x^2)/b^2 = 1, ((4)^2)/25 + ((-2)^2)/b^2 = 1, 16/25 + 4/b^2 = 1
       
      16b2 + 100 = 25b2
      100 = 9b2

      100/9 = b2

    Then my equation is:

      y^2 / 25 + x^2 / (100/9) = 1, or y^2 / 25 + (9x^2) / 100 = 1

  • Write an equation for the ellipse having foci at (2, 0) and (2, 0)
    and eccentricity
    e = 3/4.
  • The center is between the two foci, so (h, k) = (0, 0). Since the foci are 2 units to either side of the center, then c = 2, this ellipse is wider than it is tall, and a2 will go with the x part of the equation. I know that e = c/a, so 3/4 = 2/a. Solving the proportion, I get a = 8/3, so a2 = 64/9. The equation b2 = a2 c2 gives me 64/9 4 = 64/9 36/9 = 28/9 = b2.

    Now that I have values for a2 and b2, I can create my equation:

      (9x^2) / 64 + (9y^2) / 28 = 1

<< Previous  Top  |  1 | 2 | 3 | 4  |  Return to Index  Next >>

Cite this article as:

Stapel, Elizabeth. "Conics: Ellipses: Finding the Equation from Information." Purplemath.
    Available from
http://www.purplemath.com/modules/ellipse3.htm.
    Accessed
 

 



Purplemath:
  Linking to this site
  Printing pages
  School licensing


Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
   Guidelines"

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor


This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright 2010-2012  Elizabeth Stapel   |   About   |   Terms of Use

 

 Feedback   |   Error?