# Thread: HELPfocus of parabola!!

1. ## HELPfocus of parabola!!

could someone please walk through the steps of solving this problem....?

find the focus of the conic section y=-3x^2-12x-5

2. Hello, emetty90!

Find the focus of the parabola: . $y\:=\:-3x^2-12x-5$

We must get the equation into the form: . $(x-h)^2 \:=\:4p(y-k)$

Then the vertex is at $(h,k)$ and $p$ is the directed distance from the vertex to the focus.

We have: . $3x^2 + 12x \:=\:\text{-}y - 5$

Divide by 3: . $x^2 + 4x \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3}$

Complete the square: . $x^2 + 4x {\color{red}\;+\; 4} \:=\:\text{-}\tfrac{1}{3}y - \tfrac{5}{3} {\color{red}\;+\; 4} \quad\Rightarrow\quad (x+2)^2 \:=\:\text{-}\tfrac{1}{3}y+\tfrac{7}{3}$

. . And we have: . $(x+2)^2 \;=\;\text{-}\tfrac{1}{3}(y - 7)$

The vertex is: . $V(-2,7)$ . . . and the parabola opens downward.

Since $4p = \text{-}\tfrac{1}{3} \quad\Rightarrow\quad p \:=\:\text{-}\tfrac{1}{12}$ . . . The focus is $\tfrac{1}{12}$ unit below the vertex.

Therefore, the focus is: . $\left(\text{-}2,\,7\,\text{-}\,\tfrac{1}{12}\right) \;=\;\left(\text{-}2,\frac{83}{12}\right)$