Select the correct answer from each drop-down menu. the equation of a hyperbola is -4x2 + y2 + 16x + 12y + 16 = 0. the center of the hyperbola is (2, -, 2)(0, 0), and its vertices are (-4, 2)(2, -4)(2, 6) and (-8, 2)(2, -8)(2, -6)

Step-by-step explanation:

Rearranging the equation to standard form helps answer this question.

... -4(x^2 -4x) +(y^2 +12y) = -16 . . . subtract the constant, group x-terms, y-terms

... -4(x^2 -4x +4) +(y^2 +12y +36) = -16 -4(4) +36

... -4(x -2)^2 +(y +6)^2 = 4 . . . . write as squares

... ((y +6)/2)^2 -(x -2)^2 = 1 . . . . divide by 4

Now, this is in the form ...

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

which has center (h, k) and vertices (h, k±a).

This form tells you the center is (x, y) = (2, -6), and the vertices are (2, -4) and (2, -8).

1 and 2

step-by-step explanation:

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