9.2. Parabolas http://www.ck12.org
Example A
Identify the following conic, put it into graphing form and identify its vertex, focal length(p), focus, directrix and
focal width.
2 x^2 + 16 x+y= 0
Solution: This is a parabola because they^2 coefficient is zero.
x^2 + 8 x=−^12 y
x^2 + 8 x+ 16 =−^12 y+ 16
(x+ 4 )^2 =−^12 (y− 32 )
(x+ 4 )^2 =− 4 ·^18 (y− 32 )The vertex is (-4, 32). The focal length is( p=^18. This parabola opens down which means that the focus is at
− 4 , 32 −^18 )and the directrix is horizontal aty= 32 +^18. The focal width is^12.
Example B
Sketch the following parabola and identify the important pieces of information.
(y+ 1 )^2 = 4 ·^12 ·(x+ 3 )
Solution:
The vertex is at (-3, -1). The parabola is sideways because there is ay^2 term. The parabola opens to the right
because the 4pis positive. The focal length isp=^12 which means the focus is^12 to the right of the vertex at (-2.5,
-1) and the directrix is^12 to the left of the vertex atx=− 3 .5. The focal width is 2 which is why the width of the
parabola stretches from (-2.5, 0) to (-2.5, -2).
Example C
What is the equation of a parabola that has a focus at (4, 3) and a directrix ofy=−1?
Solution: It would probably be useful to graph the information that you have in order to reason about where the
vertex is.