College Algebra, Chapter 8, 8.1, Section 8.1, Problem 14

Determine the focus, directions and focal diameter of the parabola $y^2 = 3x$. Then, sketch its graph.



The equation $y^2 = 3x$ is a parabola that opening to the right. The parabola has the form $y^2 = 4px$. So


$
\begin{equation}
\begin{aligned}

4p =& 3
\\
\\
p =& \frac{3}{4}

\end{aligned}
\end{equation}
$


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