9.2. Parabolas http://www.ck12.org
y^2 − 4 y+ 12 x− 32 = 0
Answers:
- The vertex must lie directly between the focus and the directrix, so it must be at (2, 4). The focal length is
therefore equal to 1. The parabola opens downwards.
(x− 2 )^2 =− 4 · 1 ·(y− 4 ) - The focus is in the middle of the focal width. The focus is( 6 ,^52 ). The focal width is 19 which is four times the
focal length so the focal length must be^194. The vertex must be a focal length to the left of the focus, so the vertex
is at( 6 −^194 ,^52 ). This is enough information to write the equation of the parabola.
(y− 5
2
) (^2) = 4 · 19
4 ·
(x− 6 + 19
4
)
3.y^2 − 4 y+ 12 x− 32 = 0
y^2 − 4 y=− 12 x+ 32
y^2 − 4 y+ 4 =− 12 x+ 32 + 4
(y− 2 )^2 =− 12 (x− 3 )
(y− 2 )^2 =− 4 · 3 ·(x− 3 )The vertex is at (3, 2). The focus is at (0, 2). The directrix is atx=6.
Practice
- What is the equation of a parabola with focus at (1, 4) and directrix aty=−2?
- What is the equation of a parabola that opens to the left with focal width from (-2, 5) to (-2, -7)?
- What is the equation of a parabola that opens to the right with vertex at (5, 4) and focal width of 12?
- What is the equation of a parabola with vertex at (1, 8) and directrix aty=12?
- What is the equation of a parabola with focus at (-2, 4) and directrix atx=4?
- What is the equation of a parabola that opens downward with a focal width from (-4, 9) to (16, 9)?
- What is the equation of a parabola that opens upward with vertex at (1, 11) and focal width of 4?
Sketch the following parabolas by putting them into graphing form and identifying important information.