College Algebra, Chapter 8, 8.1, Section 8.1, Problem 22

Determine the focus, directions and focal diameter of the parabola $\displaystyle 8x^2 + 12y = 0$. Then, sketch its graph.



The equation $\displaystyle 8x^2 + 12y = 0; x^2 = \frac{-12}{8} y; x^2 = \frac{-3}{2} y$ is a parabola that opens upward. The parabola has the form $x^2 = 4py$. So


$
\begin{equation}
\begin{aligned}

4p =& \frac{-3}{2}
\\
\\
p =& \frac{-3}{8}

\end{aligned}
\end{equation}
$


So, the focus is at $\displaystyle \left( 0, \frac{-3}{8} \right)$ and directrix $\displaystyle y = -p = \frac{3}{8}$. Also, $\displaystyle 2p = 2 \left( \frac{-3}{8} \right) = \frac{-3}{4}$, thus the endpoints of the latus rectum are $\displaystyle \left( \frac{-3}{4}, \frac{-3}{8} \right)$ and $\displaystyle \left( \frac{3}{4}, \frac{-3}{8} \right)$. The focal diameter is $\displaystyle |4p| = \left| 4 \left( \frac{-3}{8} \right) \right| = \frac{3}{2} $ units. Therefore, the graph is

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