### 11.1 Parabolas

```Parabolas
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
Standard Equation of a Parabola: (Vertex at the origin)
Equation
2
x = 4cy
Focus
(0, c)
Directrix
y = –c
Equation
2
x = -4cy
Focus
Directrix
(0, -c)
y=c
(If the x term is squared, the parabola opens up or down)
Equation
2
y = 4cx
Focus
(c, 0)
Directrix
x = –c
Equation
2
y = -4cx
Focus
Directrix
(-c, 0)
x=c
(If the y term is squared, the parabola opens left or right)
Example 1: Determine the focus and directrix of the
parabola y = 4x2 :
x2 = 4cy
y = 4x2
4 4
x2 = 1/4y
4c = 1/4
c = 1/16
Focus: (0, c)
Focus: (0, 1/16)
Directrix: y = –c
Directrix: y = –1/16
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)
Focus: (–1, 0)
Directrix: x = –c
Directrix: x = 1
Let’s see what this parabola looks like...
Standard Equation of a Parabola: (Vertex at (h,k))
Equation
Focus
Directrix
2
(x-h) = 4c(y-k) (h, k + c) y = k - c
Equation
2
(x-h) = -4c(y-k)
Focus
(h, k - c)
Directrix
y=k+c
Standard Equation of a Parabola: (Vertex at(h,k))
Equation
2
(y-k) = 4c(x-h)
Focus
(h +c , k)
Directrix
x=h-c
Equation
Focus
2
(y-k) = -4c(x-h) (h - c, k)
Directrix
x=h+c
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.
Equation of Parabola in General Form
2
For (x-h) = 4c(y-k)
y = Ax2 + Bx + C
2
For (y-k) = 4c(x-h)
x = Ay2 + By + C
Example 4: Convert y = 2x2 -4x + 1 to standard form
y = 2x2 -4x + 1
y - 1 = 2(x2 -2x)
y – 1 + 2 = 2(x2 -2x + 1)
y + 1 = 2(x -1) 2
1(y + 1) = (x -1) 2
2
(x -1) 2 = 1(y + 1)
2
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)
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)
(2 - (-2))2 = 4c(8 - 6)
16 = 8c
2=c
x = 2 and y = 8
(x - h)2 = 4c(y - k)
(x - (-2))2 = 4(2)(y - 6)
(x + 2)2 = 8(y - 6)
Standard form
3.6.10
Example 7: Sketch (y-2)2 ≤ 12(x-3)
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.
y2 - 8x - 2y - 15 = 0
y2 - 2y + _____
1 = 8x + 15 + _____
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.
4c = 8
c=2
Example 9: Find the intersection point(s), if any, of the parabola
2
2
x
y

1
with equation y2 = 2x + 12 and the ellipse with equation
36 64
64x  36y  2304
64x 2  36(2x  12)  2304
64x 2  72x  432  2304
64x 2  72x  1872 0
2
2
```