gisellegendelman
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pls help!!!
Find an equation in standard form for the hyperbola with vertices at (0, ±4) and asymptotes at y = ±1/3.x.
a. y squared over 144 minus x squared over 16 = 1
b. y squared over 16 minus x squared over 36 = 1
c. y squared over 16 minus x squared over 144 = 1
d. y squared over 36 minus x squared over 4 = 1

Respuesta :

jiujuan

Answer:

y^2/16  -  x^2/144 =1 is your answer.

Step-by-step explanation:

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