College Algebra & Trigonometry, 2018a

274 CHAPTER 5. CONIC SECTIONS - CIRCLE AND PARABOLA
If we are given the equation of a parabola and need to find the vertex, focus and
directrix, it is often helpful to put the equation in standard form. This usually
requires completing the square.
If we’re given the equation:
x 2 +6x − 4y +1=0
then we know that this will be either an upward or downward facing parabola.
Let’s move everything away from the x terms but leave a space to complete the
square:
x 2 +6x =4y − 1
To complete the square on this, we need add 9 to both sides so that we can rewrite
the left side as (x +3) 2 .
x 2 +6x +9=4y − 1+9
(x +3) 2 =4y +8
Then we can factor out the coefficient of the y variable (even if it doesn’t factor
evenly).
(x +3) 2 =4(y +2)
So, this is an upward facing parabola with the vertex at the point (−3, −2). To
find the focus and directrix, we need to know the vlaue of p. Since 4p =4, then
we know that p =1. This means that the focus will be 1 unit above the vertex at
the point (−3, −1) and the directrix will be one unit below the vertex at the line
y = −3.
Vertex: (−3, −2)
Focus: (−3, −1)
Directrix:
y = −3