Question

Which is the equation of a hyperbola with directrices at y = ±2 and foci at (0, 4) and (0, −4)?

x squared over 8 minus y squared over 16 equals 1
y squared over 16 minus x squared over 8 equals 1
x squared over 8 minus y squared over 8 equals 1
y squared over 8 minus x squared over 8 equals 1

Answer 1

The equation of the hyperbola with directrices at y = ±2 and foci at (0, 4) and (0, −4) is y²/16 - x²/12 = 1.

Step-by-step explanation:

Given the directrices of the hyperbola at y = ±2 and the foci at (0, 4) and (0, −4), we can determine the general equation of the hyperbola. For a vertical hyperbola, the equation has the form (y-k)²/a² - (x-h)²/b² = 1, where (h, k) is the center of the hyperbola, and a and b are the distances from the center to the vertices and to the directrices, respectively.

Since the center of the hyperbola is at the origin (0, 0) due to the symmetry of the foci, and the distance between the foci (2c) is 8, this indicates that c = 4. The distance from the center to a directrix is a, which in this case is 2.

Thus, a and c satisfy the equation c² = a² + b² or 4² = 2² + b², giving b² = 12 after solving.

Therefore, the equation of the hyperbola is y²/16 - x²/12 = 1.

Answer 2

1

Step-by-step explanation:

x2/8 - y2/8 = 1

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