Michigan State University - Diagnostic Exam

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5.2

  •   Find the center and type of the conic section with equation 4x2 - 9y2 + 54y = 117.

    a) (0, 3), ellipse b) (0, 3), hyperbola c) (0, - 3), hyperbola
       
    d) (2, 3) ellipse e) (4, 9) hyperbola

Solution:

2 2 2 2 4x - 9y + 54y = 117 ⇒ 4x - 9(y - 6y + 9) = 117 - 81 = 4x2 - 9(y - 3)2 = 36

So the conic section is a hyperbola with center (0, 3) .

Diagnostic:

If you miss this question please review hyperbolas.

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July 15, 2008