Determine the center, vertices, foci and asymptotres of the hyperbola $\displaystyle \frac{x^2}{49} - \frac{y^2}{32} = 1$. Then, sketch its graph
The hyperbola has the form $\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with center at origin and horizontal transverse axis since the denominator
of $x^2$ is positive. This gives $a^2 = 49$ and $b^2 = 32$, so $a = 7, b = 4\sqrt{2}$ and $c = \sqrt{a^2 + b^2} = \sqrt{49+32} = 9$
Then, the following are determined as
$
\begin{equation}
\begin{aligned}
\text{center } (h,k) && \rightarrow && (0,0)\\
\\
\text{vertices } (\pm a,0)&& \rightarrow && (\pm 7,0)\\
\\
\text{foci } (\pm c, 0) && \rightarrow && (\pm9, 0)\\
\\
\text{asymptote } y = \pm \frac{b}{a}x && \rightarrow && y = \pm \frac{4\sqrt{2}}{7}x
\end{aligned}
\end{equation}
$
Therefore, the graph is
Thursday, June 13, 2013
College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 18
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