Question

dentify the equations of the asymptotes of the hyperbola (y - 7)2 - 16(x + 1)2 = 64.

Answer 1

\(\bf \cfrac{(y-h)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1\\ \quad \\ \quad \\
(y - 7)^2 - 16(x + 1)^2 = 64\implies \cfrac{(y-7)^2}{64}-\cfrac{(x-(-1))^2}{64}=1\\ \quad \\
\implies \cfrac{(y-7)^2}{8^2}-\cfrac{(x-(-1))^2}{8^2}=1\\ \quad \\ \quad \\
\textit{asymptotes are at }\quad y=k\pm\cfrac{a}{b}(x-h)\)

Answer 2

hmm I kinda ate the 16...

Answer 3

oh

Answer 4

thanks for helping

Answer 5

\(\bf \cfrac{(y-h)^2}{a^2}-\cfrac{(x-h)^2}{b^2}=1\\ \quad \\ \quad \\
(y - 7)^2 - 16(x + 1)^2 = 64\implies \cfrac{(y-7)^2}{64}-\cfrac{16(x-(-1))^2}{64}=1\\ \quad \\
\implies \cfrac{(y-7)^2}{8^2}-\cfrac{(x-(-1))^2}{2^2}=1\\ \quad \\ \quad \\
\textit{asymptotes are at }\quad y=k\pm\cfrac{a}{b}(x-h)\)

Answer 6

arg... yet another typo =) one sec

Answer 7

k

Answer 8

\(\bf \cfrac{(y-\color{red}{k})^2}{\color{red}{a}^2}-\cfrac{(x-\color{red}{h})^2}{\color{red}{b}^2}=1\\ \quad \\ \quad \\
(y - 7)^2 - 16(x + 1)^2 = 64\implies \cfrac{(y-7)^2}{64}-\cfrac{16(x-(-1))^2}{64}=1\\ \quad \\
\implies \cfrac{(y-7)^2}{8^2}-\cfrac{(x-(-1))^2}{2^2}=1\\ \quad \\ \quad \\
\textit{asymptotes are at }\quad y=k\pm\cfrac{a}{b}(x-h)\)

Answer 9

anyhow there

Answer 10

so just plugin for the aysmptotes right

Answer 11

pretty much, yes