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What type of conic is each?

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Hyperbolas 5.4 (M3)

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What is the equation?

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What is the equation of each graph? a. b. c.

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What is the equation?

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Place the equation with the graph. 1. 6.5.4. 3.2.

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What do you notice? Standard Formula of a hyperbola is like an ellipse with subtraction. Standard Formula of a hyperbola is like an ellipse with subtraction. Its appearance is a broken ellipse turned in the opposite direction. Its appearance is a broken ellipse turned in the opposite direction. There are 2 asymptotes. There are 2 asymptotes. Positive term is the direction the “parabola” opens. Positive term is the direction the “parabola” opens. The term subtracted becomes the axis you “cut” from the ellipse. The term subtracted becomes the axis you “cut” from the ellipse.

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Hyperbola Examples

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Hyperbola Notes Horizontal Transverse Axis Center (0,0) Vertices (a,0) & (-a,0) (-a,0) Foci (c,0) & (-c, 0) (-c, 0) Asymptotes

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Hyperbola Notes Horizontal Transverse Axis Equation

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To find asymptotes

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Hyperbola Notes Vertical Transverse Axis Center (0,0) Vertices (a,0) & (-a,0) (-a,0) Foci (c,0) & (-c, 0) (-c, 0) Asymptotes

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Hyperbola Notes Vertical Transverse Axis Equation

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To find asymptotes

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Graphing a Hyperbola 1. Graph the center. 2. Take the square root of the positive denominator to find the vertices. 3. Graph the vertices with a point. 4. Take the square root of the negative denominator to find the co-vertices. 5. Graph the co-vertices with a dash. 6. Create the box. 7. Draw the 2 asymptotes as the 2 diagonals. 8. Go back to the vertices & graph the hyperbola going towards the asymptotes. 9. To find the foci, c 2 =a 2 + b 2

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Write an equation of the hyperbola with foci (-5,0) & (5,0) and vertices (-3,0) & (3,0) a = 3 c = 5

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Write an equation of the hyperbola with foci (0,-6) & (0,6) and vertices (0,-4) & (0,4) a = 4 c = 6

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EXAMPLE 1 Graph an equation of a hyperbola Graph 25y 2 – 4x 2 = 100. Identify the vertices, foci, and asymptotes of the hyperbola. SOLUTION STEP 1 Rewrite the equation in standard form. 25y 2 – 4x 2 = 100 Write original equation. 25y 2 100 – 4x 2 100 = Divide each side by 100. y 2 4 – y 2 25 = 1 Simplify.

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EXAMPLE 1 Graph an equation of a hyperbola STEP 2 Identify the vertices, foci, and asymptotes. Note that a 2 = 4 and b 2 = 25, so a = 2 and b = 5. The y 2 - term is positive, so the transverse axis is vertical and the vertices are at (0, +2). Find the foci. c 2 = a 2 – b 2 = 2 2 – 5 2 = 29. so c = 29. The foci are at ( 0, + ) 29. (0, + 5.4). The asymptotes are y = abab + x or 2525 + x y =

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EXAMPLE 1 Graph an equation of a hyperbola STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 4 units high and 2b = 10 units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

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EXAMPLE 2 Write an equation of a hyperbola Write an equation of the hyperbola with foci at (–4, 0) and (4, 0) and vertices at (–3, 0) and (3, 0). SOLUTION The foci and vertices lie on the x -axis equidistant from the origin, so the transverse axis is horizontal and the center is the origin. The foci are each 4 units from the center, so c = 4. The vertices are each 3 units from the center, so a = 3.

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EXAMPLE 2 Write an equation of a hyperbola Because c 2 = a 2 + b 2, you have b 2 = c 2 – a 2. Find b 2. b 2 = c 2 – a 2 = 4 2 – 3 2 = 7 Because the transverse axis is horizontal, the standard form of the equation is as follows: x 2 3 2 – y27y27 = 1 Substitute 3 for a and 7 for b 2. x 2 9 – y27y27 = 1 Simplify

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GUIDED PRACTICE for Examples 1 and 2 Graph the equation. Identify the vertices, foci, and asymptotes of the hyperbola. 1. x 2 16 – y 2 49 = 1 SOLUTION (+4, 0), ( + ), 65, 0 7474 + x y =

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GUIDED PRACTICE for Examples 1 and 2 2. y 2 36 – x 2 = 1 SOLUTION (0, +6), ( 0, + ), 37 y = +6x

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GUIDED PRACTICE for Examples 1 and 2 3. 4y 2 – 9x 2 = 36 SOLUTION ( 0, + ), 13 (0, +3), 3232 + x y =

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GUIDED PRACTICE for Examples 1 and 2 Write an equation of the hyperbola with the given foci and vertices. 4. Foci: (–3, 0), (3, 0) Vertices: (–1, 0), (1, 0) SOLUTION x 2 – y28y28 = 1 5. Foci: (0, – 10), (0, 10) Vertices: (0, – 6), (0, 6) SOLUTION = 1 y 2 36 – x 2 64

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