College Algebra, Chapter 8, 8.1, Section 8.1, Problem 18

Determine the focus, directions and focal diameter of the parabola $\displaystyle x = \frac{1}{2} y^2$. Then, sketch its graph.



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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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