Question

Given the focus and directrix shown on the graph what is the vertex form of the equation of the parabola?

Answer 1

The equation of the directrix is x = -4 and the focus coordinates are (1, 3).

Since the directrix is a vertical line, we can use the model below for a parabola that opens to the right or left:

$(y-k)^2=4p(x-h)$

Where the focus is located at (h + p, k) and the directrix is x = h - p.

Since the directrix is x = -4 and the focus is (1, 3), we have:

$\begin{gathered} h+p=1 \\ h-p=-4 \\ k=3 \end{gathered}$

Adding the first two equations, we have:

$\begin{gathered} 2h=-3 \\ h=-\frac{3}{2} \\ \\ h+p=1 \\ -\frac{3}{2}+p=1 \\ p=\frac{5}{2} \end{gathered}$

Therefore the equation is:

$(y-3)^2=10(x+\frac{3}{2})$

Now, let's rewrite it in the vertex form:

$\begin{gathered} vertex\text{ }form\to x=a(y-k)^2+h \\ \\ (y-3)^2=10(x+\frac{3}{2}) \\ \frac{1}{10}(y-3)^2=x+\frac{3}{2} \\ x=\frac{1}{10}(y-3)^2-\frac{3}{2} \end{gathered}$

Therefore the correct option is A.