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Chapter 13
Conic Sections
Conic Sections
● Conic Section is a section about cone shape figures. The figures we will be learning are– Circle
– Parabola
– Ellipse
– Hyperbola
Circle
● A CIRCLE is comprised of a set of points (x, y) that have the same distance (called the radius) from the fixed point (called the center)
Circle
● Basic Form– The equation of a circle, centered at (0, 0) with a
radius r, is given by the equation
– x² + y² = r²
● Complex Form– The equation of a circle, centered at (h, k) with a
radius r, is given by the equation
– (x - h)² + (y - k)² = r²
Circle
● Given the equation of a circle determine the center and the radius (x + 2)² + y² = 25
Circle
● Graph the circle (x + 2)² + y² = 25
Parabolas
● Basic Form– The equation of a parabola centered at the origin
(0,0) (called the vertex) is given by y = ±ax²
● Complex Form– The equation of a parabola centered (h, k) (called
the vertex) is given by y = ±a(x - h)² + k
Parabola
● State the vertex of the parabola given the equation y = -2 (x – 4)² - 1
Parabola
● Graph the parabola y = -2 (x – 4)² - 1
Ellipse
● An ELLIPSE is a figure that has a horizontal distance and a vertical distance from the the center. The longer distance is the major axis and the shorter distance is the minor axis.
Ellipse
● Basic Form– The equation of an ellipse, centered at (0, 0) is
given by the equation
–
● Complex Form– The equation of an ellipse, centered at (h, k) is
given by the equation
–
x2
a2 +y2
b2 =1
(x−h)2
a2 +( y−k )2
b2 =1
Ellipse
● Graph (x+5)
2
4+( y−2)
2
16=1
Hyperbola
● A HYPERBOLA looks like a pair of parabolas, but they are not. Every hyperbola has two vertices and the line through the vertices is known as the transverse axis. The point halfway between the vertices is called the center. There are also asymptotes that cross the center.
Hyperbola
● Basic Form– The equation of a hyperbola, centered at (0, 0) with
asymptotes at and is
x2
a2−y2
b2=1
y=(ba)x y=−(ba)x
Hyperbola
● Complex Form– The equation of a hyperbola, centered at (h, k) with
asymptotes at and is
(x−h)2
a2 −( y−k )2
b2 =1
y−k=(ba)(x−h) y−k=−(ba)(x−h)
Hyperbola
● Graph and find the asymptotes y2
4−x2
16=1
Determine and Graph
● Given the polynomial determine the type of graph and graph the conic.
● x² – 4x + 4y² + 8y - 8 = 0
Homework
● Section 13.1 # 45, 47, 55● Section 13.2 # 9, 21, 27, 31● Section 13.3 # 7, 9, 59