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CHAPTER 6
College Algebra
Conic Sections
• A conic section is formed when a right circular cone with two parts,
called nappes is intersected by a plane. One of four types of curves
can be formed; a parabola, a circle, an ellipse, or a hyperbola.
6.1 Parabolas
• A parabola is the set of all points equidistant from a fixed line (the
directrix) and a fixed point not on the line (the focus).
• The line that is perpendicular to the directrix and contains the focus is the
axis of symmetry. The vertex is the midpoint of the segment between the
focus and the directrix.
• The standard equation of a parabola with vertex (0,0) and directrix
y = -p is
• x2 = 4py
• The focus is (0,p) and the y-axis is the axis of symmetry.
• The standard equation of a parabola with vertex (0,0) and directrix
x = -p is
• y2 = 4px
• The focus is (p,0) and th x-axis is the axis of symmetry.
Standard Equation for a Parabola
• Parabola with vertex (h,k) and vertical axis of symmetry
• (x - h)2 = 4p(y - k)
• Where the vertex is (h,k), the focus is (h, k +p) and the directrix is y = k –p
• Parabola with vertex (h,k) and horizontal axis of symmetry
• (y – k)2 = 4p(x – h)
• Where the vertex is (h,k) and the focus (h+p, k) and the directrix is x = h - p
If an equation is not given to you in standard
form --• Example: y2 – 2y – 8x – 31 = 0
First we complete the square (add (b/2) 2 to both sides):
• y2 – 2y + 1 = 8x + 31 + 1
(y – 1)2 = 8 (x + 4)
(y – 1)2 = 4 (2) [x – (- 4)]
So the vertex (h,k) = (-4, 1)
Focus (h + p, k) = (-4 + 2, 1) = (-2, 1)
Directrix x = h – p so x = -4 – 2 or x = - 6
6.2 The Circle and The Ellipse
• A circle is the set of all points in a plane that are at a fixed distance
from a fixed point (the center) in the plane.
• The standard equation of a circle with center (h,k) and radius r is
(x-h)2 + (y-k)2 = r2
• As we saw the parabola – sometimes the equation may not be given
to you in standard form. When that is the case you need to
complete the square (just like before):
x2 + y2 – 16x + 14y + 32 = 0
x2 – 16x + 64 + y2 + 14y + 49 = -32 + 64 + 49
(x-8)2 + (y – (-7))2 = 92
Ellipse
• An ellipse is the set of all points in a plane, the sum of whose
distances from two fixed points (the foci) is constant. The center of
an ellipse is the midpoint of the segment between the foci.
• Standard Equation of Ellipses
• Major Axis is Horizontal
(x - h)2 (y - k)2
+
=1, a > b > 0
2
2
a
b
Vertices: (h – a, k) and (h + a, k)
Length of minor axis: 2b
Foci (h –c, k) and (h+c, k)
where c2 = a2 –b2
Ellipse
• Standard equation of an Ellipse with center (h,k)
Major axis is Vertical
(x - h)2 (y - k)2
+
=1, a > b > 0
2
2
b
a
Vertices (h, k – a) and (h, k +a)
Length of minor axis: 2b
Foci (h, k –c) and (h, k + c)
where c2 = a2 – b2
Ellipse Example
• 4x2 + y2 + 24x – 2y + 21 = 0
Graph the ellipse
•
4(x2 + 6x +9) + (y2 – 2y + 1) = -21 + 36 + 1
4 (x + 3)2 + (y – 1)2 = 16
(x + 3)2 (y -1)2 16
+
=
4
16
16
Center (-3, 1) a = 4 and b = 2
(major axis vertical)
so…
(x - (-3))2 (y -1)2
+
=1
2
2
2
4
Homework
6.3 The Hyperbola
• A hyperbola is the set of all points in a given plane for which the
absolute value of the difference of the distances from two fixed
points (the foci) is constant. The midpoint of the segment between
the foci is the center of the hyperbola.
• Your book calls the conjugate axis, the transverse axis.
Standard Equation of Hyperbola with Center
(h,k)
• Transverse Axis is Horizontal:
(x - h)2 (y - k)2
=1
2
2
a
b
Vertices: (h – a, k) and (h + a, k)
Asymptotes: y – k = (b/a)(x – h) and y – k = (-b/a)(x-h)
Foci: (h – c, k) and (h + c, k) where c 2 = a2 + b2
• Transverse Axis is Vertical
(y - k)2 (x - h)2
=1
2
2
a
b
Vertices (h, k –a) and (h, k + a)
Asymptotes: y – k = (a/b)(x – h) and y – k = (-a/b)(x – h)
Foci: (h, k – c) and (h, k + c) where c 2 = a2 + b2
6.4 Nonlinear Systems of Equations and
Inequalities
• We solve nonlinear systems of equations the same way we solved linear
systems. We can graph, use substitution and occasionally use elimination.
• Example -- Solve the following system of equations:
x2 + y2 = 25 and 3x – 4y = 0
Using substitution: 3x = 4y so x = (4/3)y
((4/3)y)2 + y2 = 25
(16/9)y2 + y2 = 25
(25/9) y2 = 25
y2 = 9
y = -3 and 3
To get the x-value we can plug those y-values into any of our original
equations:
x = (4/3)3 so x = 4 when y = 3
(4,3)
x = (4/3)(-3) so x = -4 when y = -3
(-4,-3)
Nonlinear System of Inequalities
• We can solve nonlinear system of inequalities by graphing.
• Where the shaded regions overlap is solution set.
• x2 + y2 < 4
y > x2 + 1
Homework