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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.

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      (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

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Cite this article as:

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

 

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