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

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

imstuck · Answer

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?

imstuck · Answer

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:$$y=\pm \frac{ 4 }{ 3 }x$$

imstuck · Answer

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