College Algebra, Chapter 8, 8.1, Section 8.1, Problem 12

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



The equation $x^2 = y$ is a parabola that opens upward. The parabola has the form $x^2 = 4py$. So


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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