Introduction to Conic SectionsConic sections will be defined in two different ways in this unit.1. The set of points formed by the intersection

of a plane and a double-napped cone.2. The set of points satisfying certain

conditions in relationship to a fixed point and a fixed line or to two fixed points.

Conic sections are the shapes formed on a plane when that plane intersects two cones (vertex to vertex). We will discuss four different conic sections: circles, parabolas, ellipses, and hyperbolas. These four conic sections can degenerate into degenerate conic sections. The intersections of the double-napped cone and the plane are a point, a line, and intersecting lines.

http://youtu.be/GDHNoQHQmtQ

Section 10.2

Parabolas

1st Definition of a ParabolaA parabola is a conic section formed when a plane intersects one of the cones and is parallel to a diagonal side (generator) of the cone.The degenerate conic section associated with a parabola is a line.

2nd Definition of a ParabolaA parabola is a set of points in a plane that are the same distance from a given point, called the focus and a given line called directrix.

Draw a line through the focus perpendicular to the directrix. This line is the axis of the parabola. Find the point on the axis that is equidistant from the focus and the directrix of the parabola. This is the vertex of what will become a parabola. We call the distance from the focus to the vertex the focal length.

directrix

focusvertex

focal length

focal length

axis

In general, the graph of a parabola is bowl-shaped. The focus is within the bowl. The directrix is outside the bowl and perpendicular to the axis of the parabola.

F

directrix

axis

General Equation of a Parabola

Vertical AxisAx2 + Dx + Ey + F = 0

Horizontal AxisCy2 + Dx + Ey + F = 0

To rewrite from the general form to other forms you will complete the square.

Standard Equation of a ParabolaIf p = the focal length, then the standard form of the equation of a parabola with vertex at (h, k) is as follows: Vertical Axis(x – h)2 = 4p(y – k)Horizontal Axis(y – k)2 = 4p(x – h)4p = focal width: the length of the perpendicular segment through the focus whose endpoints are on the parabola.

F

directrix

axis

Focal width

Vertex Equation of a ParabolaIf p = the focal length and (h, k) is the vertex of a parabola, then the vertex form of the equation of a parabola is Vertical Axisy = a(x – h)2 + kHorizontal Axisx = a(y – k)2 + h

where 1 .4

ap

Example 1

For each parabola state the form of the given equation, horizontal or vertical, find the vertex, axis, the focal length, focus, directrix, and focal width. Graph the ones indicated.

1. 4(x − 2) = (y + 3)2 Graph.

  form:vertex:

  axis:focal length:focus:directrix:focal width:

(2, −3)Standard and horizontal

x

y

V F

4p = 4 so p = 1

4(x − 2) = (y + 3)2

1. 4(x − 2) = (y + 3)2 Graph.

  form:vertex:

  axis:focal length:focus:directrix:focal width:

(2, −3)Standard and horizontal

y = −34p = 4, p = 1

(3, −3)x = 1

4p = 4

x

y

VF

2. 2x2 + 4x – y − 3 = 0  form:

y + 3 = 2x2 + 4xy + 3 + __ = 2(x2 + 2x + __ )y + 3+ 2 = 2(x2 + 2x + 1)y + 5 = 2(x + 1)2

y = 2(x + 1)2 − 5 (vertex form)

General and vertical

vertex:axis:

focus:

directrix:

(−1, −5)x = −1

124 p

focal length:

18

p

1 71, 5 1, 48 8

1 15 or 58 8

y y

y = 2(x + 1)2 − 5

focal width: 1 14 48 2

p

3. x2 + 2y − 6x + 8 = 0 Graph.  Form:

2y + 8 = −x2 + 6x2y + 8 − 9 = −(x2 − 6x + 9)2y − 1 = −(x − 3)2

General and Vertical

2 13 2 standard form2

x y

vertex:

axis:

focal length: focal width:

focus:

directrix:

13,2

x = 3

12

p 4p = 2

1 13, 3,02 2

1 1 or 12 2

y y

2 13 22

x y

V

F

Graph.

Example 2

Write the equation for each parabola.

1. Vertex (2, 4); Focus (2, 6) in standard formp = 2

  vertical parabola4p = 8(x – 2)2 = 8(y – 4)

2. Focus (−2, 0); Directrix: x = 4 in vertex form2p = 6 so p = 3horizontal parabolaVertex: (−2 + 3, 0) = (1, 0)

21 112

x y

14

ap

14 3

1

12

3. Vertex (4, 3); Parabola passes through (5, 2) and has a vertical axis. Write in standard form.(x – 4)2 = 4p(y – 3)(5 – 4)2 = 4p(2 – 3)1 = −4p−1 = 4p(x − 4)2 = −(y – 3)