1. Parabolas

2. The parabola:
A curve on which all points are equidistant from a focus and a line called the directrix.
dPF = dPdirectrix = c or dPF + dPdirectrix = 2c

3. Standard Equation of a Parabola: (Vertex at the origin)
Equation Focus Directrix
x2 = 4cy
(0, c)
y = –c
Equation Focus Directrix
x2 = -4cy
(0, -c)
y = c
(If the x term is squared, the parabola opens up or down)

4. Equation Focus Directrix y2 = 4cx
x = –c
Equation Focus Directrix
y2 = -4cx
(-c, 0)
x = c
(If the y term is squared, the parabola opens left or right)

5. Example 1: Determine the focus and directrix of the parabola y = 4x2 :
x2 = 4cy
y = 4x
x2 = 1/4y
4c = 1/4
c = 1/16
Focus: (0, c) Directrix: y = –c
Focus: (0, 1/16) Directrix: y = –1/16

6. Let’s see what this parabola looks like...
Example 2: Graph and determine the focus and directrix of the parabola –3y2 – 12x = 0 :
–3y2 = 12x
–3y2 = 12x –3 –3
y2 = –4x
y2 = 4cx
4c = –4
c = –1
Focus: (c, 0) Directrix: x = –c
Focus: (–1, 0) Directrix: x = 1
Let’s see what this parabola looks like...

7. Standard Equation of a Parabola: (Vertex at (h,k))
Equation Focus Directrix
(x-h)2 = 4c(y-k)
(h, k + c)
y = k - c
Equation Focus Directrix
(x-h)2 = -4c(y-k)
(h, k - c)
y = k + c

8. Standard Equation of a Parabola: (Vertex at(h,k))
Equation Focus Directrix
(y-k)2 = 4c(x-h)
(h +c , k)
x = h - c
Equation Focus Directrix
(y-k)2 = -4c(x-h)
(h - c, k)
x = h + c

9. Example 3: The coordinates of vertex of a parabola are V(3, 2) and equation of directrix is y = –2. Find the coordinates of the focus.

10. Equation of Parabola in General Form
For (x-h)2 = 4c(y-k) y = Ax2 + Bx + C For (y-k)2 = 4c(x-h) x = Ay2 + By + C

11. Example 4: Convert y = 2x2 -4x + 1 to standard form
(x -1) 2 = 1(y + 1) 2

12. Example 5: Determine the equation of the parabola with a focus at
(3, 5) and the directrix at x = 9
The distance from the focus to the directrix is 6 units,
so, 2c = -6, c = -3. V(6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4c(x - h)
h = 6 and k = 5
(y - 5)2 = 4(-3)(x - 6)
(y - 5)2 = -12(x - 6)
(6, 5)

13. Example 6: Find the equation of the parabola that has a minimum at
(-2, 6) and passes through the point (2, 8).
The vertex is (-2, 6), h = -2 and k = 6.
(x - h)2 = 4c(y - k)
x = 2 and y = 8
(2 - (-2))2 = 4c(8 - 6)
16 = 8c
2 = c
(x - h)2 = 4c(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
Standard form
3.6.10

14. Example 7: Sketch (y-2)2 ≤ 12(x-3)

15. The equation of the directrix is x + 4 = 0.
Example 8: Find the coordinates of the vertex and focus, the equation of the directrix, the axis of symmetry, and the direction of opening of y2 - 8x - 2y - 15 = 0.
4c = 8
c = 2
y2 - 8x - 2y - 15 = 0
y2 - 2y + _____ = 8x _____ 1 1
(y - 1)2 = 8x + 16
(y - 1)2 = 8(x + 2)
Standard
form
The vertex is (-2, 1).
The focus is (0, 1).
The equation of the directrix is x + 4 = 0.
The axis of symmetry is y - 1 = 0.
The parabola opens to the right.

16. Example 9: Find the intersection point(s), if any, of the parabola
with equation y2 = 2x + 12 and the ellipse with equation