conic sections
TRANSCRIPT
Conics
Conic sections are plane curves that can be formed by cutting a double right circular cone with a plane at various angles.
DEFINITION
DOUBLE RIGHT CIRCULAR CONE
A circle is formed when the plane intersects one cone and is perpendicular to the axis
AXIS
An ellipse is formed when the plane intersects one cone and is NOT perpendicular to the axis.
A parabola is formed when the plane intersects one cone and is parallel to the edge of the cone.
A hyperbola is formed when the plane intersects both cones.
DEGENERATE CONIC
In analytic geometry, a conic may be defined as a plane algebraic curve of degree 2.
It can be defined as the locus of points whose distances are in a fixed ratio to some point, called a focus, and some line, called a directrix.
GENERAL EQUATION OF CONICS
π¨ππ+π©ππ+πͺππ+π«π+π¬π+π=π
π©πβπ π¨πͺ<ππ©πβπ π¨πͺ=ππ©πβπ π¨πͺ>π
Ellipse
Parabola
Hyperbol
a
DISCRIMINANT
Parabola: A = 0 or C = 0 Circle: A = C Ellipse: A = B, but both have the
same sign Hyperbola: A and C have Different
signs
The parabola is a set of points which are equidistant from a fixed point (the focus) and the fixed line (the directrix).
The Parabola
PROPERTIES
The line through the focus perpendicular to the directrix is called the axis of symmetry or simply the axis of the curve.
The point where the axis intersects the curve is the vertex of the parabola. The vertex (denoted by V) is a point midway between the focus and directrix.
The undirected distance from V to F is a positive number denoted by |a|.
The line through F perpendicular to the axis is called the latus rectum whose length is |4a|. The endpoints are and. This determines how the wide the parabola opens.
The line parallel to the latus rectum is called the directrix.
π· (π , π )
Axis of Symmetry
DirectrixLatus Rectum \4a\
VertexFocus
π³π
π³π
|a|
TYPES OF PARABOLA
ππππππππ :π π=ππππππππππ :πππππ πππππππππ :π π½ (π ,π)
π (π ,π)π³π(π ,βππ)π³π(π ,ππ)
π« : π=βπ
TYPE 1
ππππππππ :π π=βππππππππππ :πππππ ππππππππ :π π½ (π ,π)
π (βπ ,π)π³π(βπ ,βππ)π³π(βπ ,ππ)
π« : π=π
TYPE 2
2L
)2,( aa
ππππππππ :ππ=ππππππππππ :ππππππ ππππ :π π½ (π ,π)
π (π ,π) π³π(βππ ,π)π³π(ππ ,π)
π« : π=βπ
TYPE 3
ππππππππ :ππ=βππππππππππ :π πππππππ ππππ :π π½ (π ,π)
π (π ,βπ )π³π(ππ ,βπ)π³π(βππ ,βπ)
π« : π=π
TYPE 4
Sample Problem
1.Locate the coordinates of the vertex (V), focus (F), endpoints of the latus rectum (), the equation of the directrix, and sketch the graph of .
solution1. takes the form 2. the parabola opens downward3. Compute the value of 4. so, , or 5. the required coordinates areπ½ (π ,π)π (π ,βπ)=π (π ,βπ)
π³π (ππ ,βπ )=π³π(π ,βπ)π³π (βππ ,βπ )=π³π(βπ ,βπ) π« : π=π
π« : π=π
π³π(βπ ,βπ)
π
π³π(π ,βπ)
π=π
π (π ,βπ)
π½ (π ,π)
π
| | | | | 1 2 3 4 5
| | | | | -5 -4 -3 -2 -1
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3
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-3
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2
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Sketch the graphs and determine the coordinates of V, F, ends of LR, and equation of the directrix.
1. 2. 3.