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+(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+(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=(1)/(12) \\ \text{and} \\ k=2 \end{gathered}`

by substituting these values into the above equation:

`5=k+(1)/(4a)`

we have

`5=2+(1)/(4((1)/(12)))`

which gives

`\begin{gathered} 5=2+(1)/((1)/(3))=2+3 \\ \text{then} \\ 5=5 \\ \text{which corroborate the result.} \end{gathered}`

Therefore, the answer is the last option:

`y=(1)/(12)(x+2)^2+2`