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15 September, 01:46

Find the center, vertices, and foci of the ellipse with equation 3x2 + 8y2 = 24

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Answers (2)
  1. 15 September, 01:51
    0
    Centre: (0,0)

    Vertices: (2sqrt (2), 0) (-2sqrt (2), 0)

    Co-vertices: (0, sqrt (3)) (0, - sqrt (3))

    Foci: (sqrt (5), 0) (-sqrt (5), 0)

    Step-by-step explanation:

    (x - h) ²/a² + (y - k) ²/b² = 1

    3x²/24 + 8y²/24 = 1

    x²/8 + y²/3 = 1

    Centre: (0,0)

    Vertices:

    y² = 3

    y = + / - sqrt (3)

    (0, sqrt (3))

    (0, - sqrt (3))

    x² = 8

    x = + / - sqrt (8) = + / - 2sqrt (2)

    (2sqrt (2), 0)

    (-2sqrt (2), 0)

    Foci: (c, 0)

    c² = a² - b²

    c² = 8 - 3 = 5

    c = + / - sqrt (5)
  2. 15 September, 02:07
    0
    The center is at (0,0)

    The vertices are at ((±2 sqrt (2),0)

    foci are (±sqrt (5),0)

    Step-by-step explanation:

    3x^2 + 8y^2 = 24

    Divide each side by 24

    3x^2 / 24 + 8y^2/24 = 24/24

    x^2/8 + y^2 / 3 = 1

    The general equation of an ellipse is

    (x-h) ^2 / a^2 + (y-k) ^2 / b^2 = 1

    a>b (h, k) is the center

    the coordinates of the vertices are (±a, 0)

    the coordinates of the foci are (±c, 0), where ^c2=a^2-b^ 2

    The center is at (0,0)

    a = sqrt (8) = 2sqrt (2)

    The vertices are at ((±2 sqrt (2),0)

    c = 8 - 3 = 5

    foci are (±sqrt (5),0)
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