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svp [43]
1 year ago
15

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 pa

rabola?

Mathematics
1 answer:
Kisachek [45]1 year ago
6 0

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

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