Label the foci, the major vertices, and the minor vertices. (𝑥−1)^2/49 +(𝑦+3)^2/169 =1?

1 Answer

Relevance
  • 9 months ago
    Favorite Answer

    The standard form of an equation for an ellipse with center (h, k) is

    [(x - h)² / a²] + [(y - k)² / b²] = 1

    Comparing that with the equation in the question,

    the center is (h, k) = (1, -3)

    a² = 49 , so

    a = √49 = 7 …………… only interested in the positive square root here

    b² = 169 , so

    b = √169 = 13

    b > a , so the major axis is parallel to the y-axis.

    (i.e., the major axis is vertical)

    For the minor axis vertices, start at the x-value of the center point (1, -3) , then go a = 7 units to both the left and right.

    minor axis vertices:

    ( (1 - 7) , -3 ) = (-6, -3)

    ( (1 + 7) , -3 ) = (8, -3)

    For the major axis vertices, start at the y-value of the center point (1, -3) , then go b = 13 units both upwards and downwards.

    major axis vertices:

    ( 1 , (-3 - 13) ) = (1, -16)

    ( 1 , (-3 + 13) ) = (1, 10)

    For an ellipse with major axis parallel to the y-axis, the foci are

    (h, k - c) and (h, k + c) where c = √(b² - a²)

    In this case,

    c = √(13² - 7²)

    = √(169 - 49)

    = √120

    = √(4 * 30)

    = √(2² * 30)

    = 2√30

    so the foci are

    (1, -3 - 2√30) and (1, -3 + 2√30)

    Attachment image
    • Login to reply the answers
Still have questions? Get your answers by asking now.