Question

Given the hyperbola (y2/16) – (x2/9) = 1, find the equations for its asymptotes.
A. x = (2/3)y, x = (–2/3)y
B. y = (2/3)x, y = (–2/3)x
C. x = (4/3)y, x = (–4/3)y
D. y = (4/3)x, y = (–4/3)x

Answer 1

The equation is set up so that the first coordinate mentioned, either the x or the y, defines the type of hyperbola you have. Each type has its own asymptote formula. You have here a y^2 hyperbola. Also, the a^2 is under the first fraction, and the b^2 is under the second fraction. if a^2 is 16, then a is 4; if b^2 is 9, then b is 3. With me so far?

Answer 2

the formula for the asymptotes for a y^2 hyperbola is y = +/-a/b. Your a is 4 and your b is 3. So the equations for the asymptotes are:\[y=\pm \frac{ 4 }{ 3 }x\]