College Algebra, Chapter 8, 8.1, Section 8.1, Problem 16

Determine the focus, directions and focal diameter of the parabola $y = -2x^2$. Then, sketch its graph.



The equation $\displaystyle y = -2x^2; x^2 = \frac{-1}{2}y$ is a parabola that opens downward. The parabola has the form $x^2 = 4py$. So


$
\begin{equation}
\begin{aligned}

4p =& \frac{-1}{2}
\\
\\
p =& \frac{-1}{8}

\end{aligned}
\end{equation}
$


So, the focus is at $\displaystyle (0,p) = \left(0, \frac{-1}{8} \right)$ and directrix $\displaystyle y = -p = \frac{-1}{8}$. Also, $\displaystyle 2p = 2 \left( \frac{-1}{8} \right) = \frac{-1}{4}$, thus the endpoints of the latus rectum are at $\displaystyle \left( \frac{-1}{4}, \frac{-1}{8} \right)$ and $\displaystyle \left( \frac{1}{4}, \frac{-1}{8} \right)$. The focal diameter is $\displaystyle |4p| = \left| 4 \left( \frac{-1}{8} \right) \right| = \frac{1}{2} $ units. Therefore, the graph is