Question

If a parabola's focus is at (−2, 5) and the directrix is at y = −1, what is the vertex form of the equation representing this parabola?

Answer 1

For the parabola equation in standard form

$y=a(x-h)^2+k$

we know that the focus coordinate is

$(h,k+\frac{1}{4a})$

Since our focus is (-2,5), by comparing this coordinate with the last result, we can see that

$\begin{gathered} h=-2 \\ 5=k+\frac{1}{4a} \end{gathered}$

Then, the possible solution has the form

$\begin{gathered} y=a(x-(-2))^2+k \\ or\text{ equivalently} \\ y=a(x+2)^2+k \end{gathered}$

We can see tha the last option has this form. Lets corroborate that this is the correct choice. Then, we have that

$\begin{gathered} a=\frac{1}{12} \\ \text{and} \\ k=2 \end{gathered}$

by substituting these values into the above equation:

$5=k+\frac{1}{4a}$

we have

$5=2+\frac{1}{4(\frac{1}{12})}$

which gives

$\begin{gathered} 5=2+\frac{1}{\frac{1}{3}}=2+3 \\ \text{then} \\ 5=5 \\ \text{which corroborate the result.} \end{gathered}$

Therefore, the answer is the last option:

$y=\frac{1}{12}(x+2)^2+2$