Question

question 19 · 1 point
identify the asymptotes of the hyperbola with equation
\\(\frac{(y + 3)^2}{4}-\frac{(x - 3)^2}{16}=1\\)
select the correct answer below:
the asymptotes are \\(y = \pm2(x - 3)+3\\).
the asymptotes are \\(y = \pm\frac{1}{2}(x + 3)-3\\).
the asymptotes are \\(y = \pm\frac{1}{2}(x - 3)+3\\).
the asymptotes are \\(y = \pm\frac{1}{2}(x - 3)-3\\).
the asymptotes are \\(y = \pm2(x + 3)-3\\).
the asymptotes are \\(y = \pm2(x - 3)-3\\).

Explanation:

Step1: Recall hyperbola asymptote formula

For a hyperbola of the form $\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1$, the equations of the asymptotes are $y - k=\pm\frac{a}{b}(x - h)$.

Step2: Identify values of $a$, $b$, $h$, $k$

In the given equation $\frac{(y + 3)^2}{4}-\frac{(x - 3)^2}{16}=1$, we have $a^2 = 4$, so $a = 2$; $b^2=16$, so $b = 4$; $h = 3$; $k=-3$.

Step3: Substitute values into formula

Substituting into $y - k=\pm\frac{a}{b}(x - h)$, we get $y+3=\pm\frac{2}{4}(x - 3)$, which simplifies to $y=\pm\frac{1}{2}(x - 3)-3$.

Answer:

The asymptotes are $y=\pm\frac{1}{2}(x - 3)-3$.