College Algebra, Chapter 8, 8.1, Section 8.1, Problem 50

Determine the equation of the parabola whose graph is shown below.







The equation $x^2 = 4py$ has vertex on the origin and opens upward. Also, it has focus at $(0, p)$. It shows that the focus is the $y$-intercept of the given line. Thus, by using slope intercept form, $y = mx + b$.

$\displaystyle y = \frac{1}{2} x + p$

And if the line pass through $(2, y)$ then


$
\begin{equation}
\begin{aligned}

y =& \frac{1}{2} x + p && \text{and} &&& x^2 = 4py
\\
\\
y =& \frac{1}{2} (2) + p && \text{and} &&& 2^2 = 4py
\\
\\
y =& 1 + p && \text{and} &&& 1 = py
\\
\\
\text{Thus } & && &&&
\\
\\
p =& y - 1 \qquad \text{Equation 1} && \text{and} &&& p =& \frac{1}{y} \qquad \text{Equation 2}

\end{aligned}
\end{equation}
$



By using equations 1 and 2


$
\begin{equation}
\begin{aligned}

y - 1 =& \frac{1}{y}
&&
\\
\\
y^2 - y =& 1
&& \text{Complete the square: add } \left( \frac{-1}{2} \right)^2 = \frac{1}{4}
\\
\\
y^2 - y + \frac{1}{4} =& 1 + \frac{1}{4}
&& \text{Perfect Square}
\\
\\
\left(y - \frac{1}{2} \right)^2 =& \frac{5}{4}
&& \text{Take the square root}
\\
\\
y - \frac{1}{2} =& \pm \sqrt{\frac{5}{4}}
&& \text{Add } \frac{1}{2}
\\
\\
y =& \frac{1}{2} \pm \frac{\sqrt{5}}{2}
&& \text{Choose } y > 0
\\
\\
y =& \frac{1 + \sqrt{5}}{2}
&&

\end{aligned}
\end{equation}
$


So if $\displaystyle y = \frac{1 + \sqrt{5}}{2}$, then $\displaystyle p = y - 1 = \frac{1 + \sqrt{5}}{2} - 1 = \frac{-1 + \sqrt{5}}{2}$. Therefore, the equation is $\displaystyle x^2 = 4py; x^2 = 4 \left( \frac{-1 + \sqrt{5}}{2} \right) y; x^2 = 2 \left( -1 + \sqrt{5} \right) y$.