algebra 2 help?

Question

Suppose you are given the formula (y+3)/4 = ± (x+2)/9 for the asymptotes of a hyperbola. What is the formula for the hyperbola?

A You can’t tell because when you square both sides you might introduce extraneous roots.
B (y+3)^2/16 – (x+2)^2/81 = 1
C (y+3)^2/4 – (x+2)^2/9 = 1
D (x+2)^2/81 – (y+3)^2/16 = 1
E (x+2)^2/9 – (y+3)^2/4 = 1
F You can’t tell because you don’t know whether it opens to the sides or up-and-down.

Answer 1

The canonical form of the hyperbola is:

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

The equations of the asymptotes derived from this form are:

y = k + (b/a)(x-h) and k - (b/a)(x-h)

Since the problem states that the asymptotes are:

(y+3)/4 = (x+2)/9 and (y+3)/4 = -(x+2)/9

transforming into the asymptote standard form yields:

(y+3)/4 = (x+2)/9

y+3 = 4(x+2)/9

y = -3 + (4/9)(x-(-2))

This directly gives a=9, b=4, k=-3, and h=-2.

(y+3)/4 = -(x+2)/9

y+3 = -4(x+2)/9

y = -3 - (4/9)(x+2)

y = -3 - (4/9)(x-(-2))

This verifies a=9, b=4, k=-3, and h=-2.

Plugging these values for a, b, h, and k into the canonical form yields:

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

[(x-(-2))/9]^2 - [(y-(-3))/4]^2 = 1

[(x+2)/9]^2 - [(y+3)/4]^2 = 1

Converting to form of the available answers gives:

(x+2)^2/81 - (y+3)^2/16 = 1

Answer is D.

Answer 2

F