Questions: Graph the given hyperbola (y-6)^2/4 - (x-3)^2/16 = 1.

Solution Steps

The equation of the hyperbola is given in the standard form: $\frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$. Here, $(h, k)$ is the center of the hyperbola. Comparing the given equation $\frac{(y-6)^{2}}{4}-\frac{(x-3)^{2}}{16}=1$ with the standard form, we have: Center $(h, k) = (3, 6)$ $a^2 = 4 \Rightarrow a = 2$ $b^2 = 16 \Rightarrow b = 4$

Since the $y$ term is positive, the hyperbola opens vertically. The vertices are $(h, k \pm a)$, so the vertices are $(3, 6 \pm 2)$, which are $(3, 8)$ and $(3, 4)$.

The asymptotes are given by the equations $y - k = \pm \frac{a}{b}(x - h)$. In our case, $y - 6 = \pm \frac{2}{4}(x - 3)$, which simplifies to $y - 6 = \pm \frac{1}{2}(x - 3)$. So, the asymptotes are $y = \frac{1}{2}(x - 3) + 6$ and $y = -\frac{1}{2}(x - 3) + 6$.

Plot the center $(3, 6)$. Plot the vertices $(3, 8)$ and $(3, 4)$. Draw the asymptotes using the equations $y = \frac{1}{2}(x - 3) + 6$ and $y = -\frac{1}{2}(x - 3) + 6$. Sketch the hyperbola branches opening vertically, passing through the vertices and approaching the asymptotes.

Final Answer