http://www.ck12.org Chapter 9. Conics
The vertex must be halfway between the focus and the directrix. This places it at (4, 1). The focal length is 2. The
parabola opens upwards. This is all the information you need to create the equation.
(x− 4 )^2 = 4 · 2 ·(y− 1 )
OR(x− 4 )^2 = 8 (y− 1 )
Concept Problem Revisited
Where should the receptor be located on a satellite dish that is four feet wide and nine inches deep?
Since real world problems do not come with a predetermined coordinate system, you can choose to make the vertex
of the parabola at (0, 0). Then, if everything is done in inches, another point on the parabola will be (24, 9). (Many
people might mistakenly believe the point(48, 9)is on the parabola but remember that half this width stretches to
(-24, 9)as well.)Using these two points, the focal width can be found.
(x− 0 )^2 = 4 p(y− 0 )
( 24 − 0 )^2 = 4 p( 9 − 0 )
242
4 · 9 =p
16 =pThe receptor should be sixteen inches away from the vertex of the parabolic dish.
Vocabulary
Thefocusof a parabola is the point that the parabola seems to curve around.
Thedirectrixof a parabola is the line that the parabola seems to curve away from.
Aparabolais the collection of points that are equidistant from a fixed focus and directrix.
Guided Practice
- What is the equation of a parabola with focus at (2, 3) and directrix aty=5?
- What is the equation of a parabola that opens to the right with focal width from (6, -7) to (6, 12)?
- Sketch the following conic by putting it into graphing form and identifying important information.