Saturday, March 2, 2019

College Algebra, Chapter 8, 8.2, Section 8.2, Problem 14

Determine the vertices, foci and eccentricity of the ellipse $4x^2 + y^2 = 16$. Determine the lengths of the major and minor
axes, and sketch the graph.
If we divide both sides by $16$, then we have
$\displaystyle \frac{x^2}{4} + \frac{y^2}{16} = 1$
We'll see that the function has the form $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$. Since the denominator of $y^2$ is larger,
than the ellipse has a vertical major axis. Thus, gives $a^2 = 16$ and $b^2 = 4$. So, $c^2 = a^2 - b^2 = 16 - 4 = 12$.
Thus, $a = 4$, $b = 2$ and $c = \sqrt{12}$. Then the following are determined as

$
\begin{equation}
\begin{aligned}
\text{Vertices}& &( \pm a,0) &\rightarrow (0, \pm 4)\\
\\
\text{Foci}& &(0, \pm c) &\rightarrow (0, \pm \sqrt{12})\\
\\
\text{Eccentricity (e)}& &\frac{c}{a} &\rightarrow \frac{\sqrt{12}}{4} \text{ or } \frac{\sqrt{3}}{2}\\
\\
\text{Length of major axis}& &2a &\rightarrow 8\\
\\
\text{Length of minor axis}& &2b &\rightarrow 4
\end{aligned}
\end{equation}
$

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