4 Edition: College Algebra & Trigonometry
4 Edition: College Algebra & Trigonometry
4 Edition: College Algebra & Trigonometry
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10.4
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Characteristics
The graphs of parabolas, circles, ellipses, and hyperbolas are called conic sections since each graph can be obtained by cutting a cone with a plane, as suggested by Figure 1 at the beginning of the chapter. All conic sections of the types presented in this chapter have equations of the general form
Ax 2 Cy 2 Dx Ey F 0,
where either A or C must be nonzero.
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Example x2 y 4 = 0 x + y2 4y = 0 x2 + y2 16 = 0
Ellipse Hyperbola
A C, AC > 0 AC < 0
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Equation
Graph
Opens up if p > 0 (or a > 0), down if p < 0 (or a < 0). Vertex is (h, k). Axis of symmetry is x = h. x2 term y is not squared.
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Description
Identification
Equation
Graph
Opens to the right if p > 0 (or a > 0), to the left if p < 0 (or a < 0). Vertex is (h, k). Axis of symmetry is y = k. y2 term x is not squared.
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Description
Identification
Equation
(x h)2 + (y k)2 = r 2
Graph
Description
Identification
Equation
x2 y 2 2 1 (a b ) 2 a b
Graph
Description
x-intercepts are a and a. y-intercepts are b and b. Horizontal major axis, length = 2a. x2 and y2 terms have different positive coefficients.
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Identification
Equation
x2 y 2 2 1 (a b ) 2 b a
Graph
Description
x-intercepts are b and b. y-intercepts are a and a. Vertical major axis, length = 2a. x2 and y2 terms have different positive coefficients.
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Identification
Equation
x2 y 2 2 1 2 a b
Graph
Description
x-intercepts are a and a. Asymptotes are found from (a, b), (a, b),( a, b), and ( a, b). x2 term has a positive coefficient. y2 term has a negative coefficient.
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Identification
Equation
y 2 x2 2 1 2 a b
Graph
Description
y-intercepts are a and a. Asymptotes are found from (b, a), (b, a),( b, a), and ( b, a). y2 term has a positive coefficient. x2 term has a negative coefficient.
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Identification
Example 1
Determine the type of conic section represented by each equation, and graph it.
a. x 2 25 5 y 2 Solution
Divide each term by 25.
x 5 y 25
2 2
Subtract 5y2.
x2 y 2 1 25 5
Divide by 25.
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Example 1
Determine the type of conic section represented by each equation, and graph it.
a. x 2 25 5 y 2 Solution
The equation represents a hyperbola centered at the origin, with asymptotes
x2 y 2 0, or 25 5
5 y x. 5
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Example 1
Determine the type of conic section represented by each equation, and graph it.
a. x 2 25 5 y 2 Solution
The x-intercepts are 5; the graph is shown here.
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Example 1
2 2
b. x 8 x y 10 y 41 Solution
2
x 2 8 x y 2 10 y 41
2
( x 2 8 x 16) 16 ( y 2 10 y 25) 25 41
Regroup terms.
( x 4) ( y 5) 41 16 25
2 2
( x 4)2 ( y 5)2 0
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Example 1
2 2
b. x 8 x y 10 y 41 Solution
The resulting equation is that of a circle with radius 0; that is, the point (4, 5). If we had obtained a negative number on the right (instead of 0), the equation would have no solution at all, and there would be no graph.
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Example 1
2 2 4 x 16 x 9 y 54 y 61 c.
Solution The coefficients of the x2 -and y2 -terms are unequal and both positive, so the equation might represent an ellipse but not a circle. (It might also represent a single point or no points at all.)
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Example 1
2 2 4 x 16 x 9 y 54 y 61 c.
Solution 2 4( x 4 x
2
) 9( y 6 y
2
) 61
4( x 4 x 4 4) 9( y 6 y 9 9) 61
2
4( x 4 x 4) 16 9( y 6 y 9) 81 61
2 2
Multiply. 4( 4) = 16
Distributive property.
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Example 1
2 2 4 x 16 x 9 y 54 y 61 c.
Solution
4( x 2) 9( y 3) 36
2 2
( x 2) ( y 3) 1 9 4
2 2
Divide by 36.
This equation represents an ellipse having center (2, 3) and graph as shown here.
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Example 1
2 x d. 6 x 8 y 7 0
Solution Since only one variable is squared (x, and not y), the equation represents a parabola. Get the term with y (the variable that is not squared) alone on one side.
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Example 1
2 x d. 6 x 8 y 7 0
Solution
8y x 6 x 7
2
8 y ( x 2 6 x
2
)7
8 y ( x 6 x 9 9) 7
Complete the square.
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Example 1
2 x d. 6 x 8 y 7 0
Solution
8 y ( x 6 x 9) 9 7
2
Distributive property; ( 9) = + 9
8 y ( x 3) 16
2
Factor; add.
1 2 y ( x 3) 2 8 1 y 2 ( x 3)2 8
Multiply by .
Subtract 2.
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Example 1
2 x d. 6 x 8 y 7 0
Solution
The parabola has vertex (3, 2) and opens down, as shown in the graph here. An equivalent form for this parabola is
( x 3) 8( y 2).
2
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Example 2
Solution
4( y 4 y
2
) 9( x 2 x
2
) 43
4( y 2 4 y 4 4) 9( x 2 2 x 1 1) 43
Complete the square.
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Example 2
Solution 2 2 4( y 4 y 4) 16 9( x 2 x 1) 9 43
Distributive property.
4( y 2)2 9( x 1)2 36
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Example 2
Solution
4( y 2)2 9( x 1)2 36
Because of the 36, we might think that this equation does not have a graph. However, the minus sign in the middle on the left shows that the graph is that of a hyperbola.
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Example 2
Solution 2 2 ( x 1) ( y 2) 1 4 9
Divide by 36; rearrange terms.
Be careful here.
This hyperbola has center (1, 2). The graph is shown here.
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The following geometric definition applies to all conic sections except circles, which have e = 0.
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is a conic section of eccentricity e. The conic section is a parabola when e = 1, an ellipse when 0 < e < 1, and a hyperbola when e > 1.
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