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Parabola Conic section
ParabolaConic section
Warm-upGraph the following parabola using:
I Finding the solution of the equations (Factoring)
II Finding the VERTEX (Using formula)
III Graphing on y-axis (using vertex)
1. y = x2 - 6x + 8
2. y = –x2 + 4x – 4
3. y = x2 - 5
Parabola: the set of points in a plane that are the same distance from a given point called the focus and a given line called the directrix.
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• Note the line through the focus, perpendicular to the directrix Axis of symmetry
• Note the point midway between the directrix and the focus Vertex
The equation of a parabola with vertex (0, 0) and focus on the y-axis is x2 = 4py.
The coordinates of the focus are (0, p).The equation of the directrix is y = -p.
If p > 0, the parabola opens up.If p < 0, the parabola opens down.
Standard equation of a
PARABOLAThe equation of a parabola with vertex (0, 0) and focus on the x-axis is y2 = 4px.
The coordinates of the focus are (p, 0).The equation of the directrix is x = -p.
If p > 0, the parabola opens right.If p < 0, the parabola opens left.
A parabola has vertex (0, 0) and the focus on an axis.Write the equation of each parabola.
Since the focus is (-6, 0), the equation of the parabola is y2 = 4px.p is equal to the distance from the vertex to the focus, therefore p = -6.
The equation of the parabola is y2 = -24x.
b) The directrix is defined by x = 5.
The equation of the parabola is y2 = -20x.
Finding the Equation of a Parabola with Vertex (0, 0)
The equation of the directrix is x = -p, therefore -p = 5 or p = -5.Since the focus is on the x-axis, the equation of the parabola is y2 = 4px.
c) The focus is (0, 3).
a) The focus is (-6, 0).
Since the focus is (0, 3), the equation of the parabola is x2 = 4py.p is equal to the distance from the vertex to the focus, therefore p = 3.
The equation of the parabola is x2 = 12y.
Practice
A parabola has vertex (0, 0) and the focus on an axis.Write the equation of each parabola.
Finding the Equation of a Parabola with Vertex (0, 0)
b) The directrix is defined by x = 3.
c) The focus is (0, -5).
a) The focus is (8, 0).
The equation of the parabola is y2 = 32x.
The equation of the parabola is y2 = -12x.
The equation of the parabola is x2 = -20y.
Finding the FOCUS DIRECTRIX
y = 4(4py)
y = 16py
1 = 16p
1/16 = p
FOCUS: (0, 1/16)
Directrix Y = - 1/16
y = 4x2 x = -3y2
x = -3(4px)
x = -12px
1 = -12p
-1/12 = pFOCUS: (-1/12, 0)Directrix x = 1/12
Practicey = 8x2 x = -4y2
FOCUS: (0, 1/32)
Directrix Y = - 1/32
FOCUS: (-1/16, 0)
Directrix x = 1/16
ParabolaConic section
WARM -UP
1. (try this 1. (try this one on your one on your own)own)
y = -6xy = -6x22
FOCUSFOCUS
(0, -(0, -11//2424))
DirectrixDirectrix
y = y = 11//2424
2. (try this 2. (try this one on your one on your own)own)
x = 8yx = 8y22
FOCUSFOCUS
(1/32, 0)(1/32, 0)
DirectrixDirectrix
x = -32x = -32
Find the focus and directrix of the following:
(y-k)2 =4p(x-h),p≠0Horizontal Axis, directrix: x = h-p
• The equation of the axis of symmetry is y = k.
• The coordinates of the focus are (h + p, k).
• The equation of the directrix is x = h - p.
Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h) h = 6 and k = 5
Standard form
y2 - 10y + 25 = -12x + 72y2 + 12x - 10y - 47 = 0 General form
(y - 5)2 = 4(-3)(x - 6)(y - 5)2 = -12(x - 6)
Write the equation of the parabola with a focus at (4, 6) and the directrix at x = 8, in standard form and general form.
Practice
(y-6)2 = -8(x-6) Standard Form
Vertex: (6,6)
y2 + 8x -12y -12 General Form
Standard Equation of a Parabola with vertex at
(h,k)(x-h)2 =4p(y-k),p≠0
Vertical Axis, directrix: y = k-p• The equation of theaxis of symmetry is x = h.
• The coordinates of the focus are (h, k + p).
• The equation of the directrix is y = k - p.
The general form of the parabola is Ax2 + Cy2 + Dx + Ey + F = 0where A = 0 or C = 0.
Find the equation of the parabola that has a min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equationand solve for p:
(x - h)2 = 4p(y - k)
(2 - (-2))2 = 4p(8 - 6) 16 = 8p 2 = p
x = 2 and y = 8
(x - h)2 = 4p(y - k)(x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form
x2 + 4x + 4 = 8y - 48x2 + 4x - 8y + 52 = 0 General form
Find the equation of the parabola that has a maximum at (3, 6) and passes through the point (9, 5).
(x-3)2 = -36(y-6) Standard Form
Vertex: (3,6)
x2 - 6x +36y -207 General Form
homework
Find the equation of the parabola that has a vertex at (2,1) and focus (2,4).
(x-h)2 = 4p (y-k)
h=2, k=1, p= 4-1 = 3
(x-2)2 = 4(3) (y-1)
(x-2)2 = 12 (y-1) Standard Form
X2 - 4x -12y + 16 = 0 General Form
• The equation of the axis of symmetry is y = k.
• The coordinates of the focus are (h + p, k).
• The equation of the directrix is x = h - p.
• The equation of theaxis of symmetry is x = h.
• The coordinates of the focus are (h, k + p).
• The equation of the directrix is y = k - p.
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 + _____ = 8x + 15 + _____1 1
(y - 1)2 = 8x + 16(y - 1)2 = 8(x + 2)
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.
4p = 8 p = 2
Standardform
Analyzing a Parabola
Finding the FOCUS DIRECTRIX
y = 4(4py)
y = 16py
1 = 16p
1/16 = p
FOCUS: (0, 1/16)
Directrix Y = - 1/16
y = 4x2 x = -3y2
x = -3(4px)
x = -12px
1 = -12p
-1/12 = pFOCUS: (-1/12, 0)Directrix x = 1/12
Find the equation of the parabola that has a min at (-2, 6) and passes through the point (2, 8).
The axis of symmetry is parallel to the y-axis.The vertex is (-2, 6), therefore, h = -2 and k = 6.
Substitute into the standard form of the equationand solve for p:
(x - h)2 = 4p(y - k)
(2 - (-2))2 = 4p(8 - 6) 16 = 8p 2 = p
x = 2 and y = 8
(x - h)2 = 4p(y - k)(x - (-2))2 = 4(2)(y - 6) (x + 2)2 = 8(y - 6) Standard form
x2 + 4x + 4 = 8y - 48x2 + 4x - 8y + 52 = 0 General form
Ex: Write the equation of the parabola with a focus at (3, 5) and the directrix at x = 9, in standard form and general form.
The distance from the focus to the directrix is 6 units, therefore, 2p = -6, p = -3. Thus, the vertex is (6, 5).
The axis of symmetry is parallel to the x-axis:
(y - k)2 = 4p(x - h) h = 6 and k = 5
Standard form
y2 - 10y + 25 = -12x + 72y2 + 12x - 10y - 47 = 0 General form
(y - 5)2 = 4(-3)(x - 6)(y - 5)2 = -12(x - 6)
