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 – 4x = Divide each side by 100. y 2 4 – y 2 25 = 1 Simplify.

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 x y =

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.

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.

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 – y27y27 = 1 Substitute 3 for a and 7 for b 2. x 2 9 – y27y27 = 1 Simplify

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 STEP 1 The equation is in standard form. x 2 16 – y 2 49 = 1

GUIDED PRACTICE for Examples 1 and 2 STEP 2 Identify the vertices, foci, and asymptotes. Note that a 2 = 16 and b 2 = 49, so a = 4 and b = 7. The x 2 - term is positive, so the transverse axis is horizontal and the vertices are at (+4, 0). Find the foci. c 2 = a 2 + b 2 = = 65. so c = The foci are at ( + ) 65, 0 The asymptotes are y = baba + x or x y =

GUIDED PRACTICE for Examples 1 and 2 STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 8 units high and 2b = 14 units wide. The asymptotes pass through opposite corners of the rectangle. Then, draw the hyperbola passing through the vertices and approaching the asymptotes.

GUIDED PRACTICE for Examples 1 and 2 2. y 2 36 – x 2 = 1 SOLUTION STEP 1 The equation is in standard form. y 2 36 – x21x21 = 1 STEP 2 Identify the vertices, foci, and asymptotes. Note that a 2 = 36 and b 2 = 1, so a = 6 and b = 1. The y 2 - term is positive, so the transverse axis is horizontal and the vertices are at (0, +6). Find the foci.

GUIDED PRACTICE for Examples 1 and 2 c 2 = a 2 + b 2 = = 37. so c = + 37 The foci are at ( 0, + ) 37 The asymptotes are y = baba + x or x y = = + 6x

GUIDED PRACTICE for Examples 1 and 2 STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 12 units high and 2b = 2 units wide. The asymptotes pass through opposite corners of the rectangle. Then draw the hyperbola passing through the vertices and approaching the asymptotes.

GUIDED PRACTICE for Examples 1 and y 2 – 9x 2 = 36 SOLUTION STEP 1 The equation is in standard form. 4y 2 – 9x 2 = 36 Write original equation. Divide each side by 36. y 2 9 – x24x24 = 1 Simplify. 4y 2 36 – 9x 2 36 = 1

GUIDED PRACTICE for Examples 1 and 2 STEP 2 Identify the vertices, foci, and asymptotes. Note that a 2 = 9 and b 2 = 4, so a = 3 and b = 2. The y 2 - term is positive, so the transverse axis is horizontal and the vertices are at (0, +3). Find the foci. c 2 = a 2 + b 2 = = 13. so c = + 13 The foci are at ( 0, + ) 13 The asymptotes are y = baba + x or x y =

GUIDED PRACTICE for Examples 1 and 2 STEP 3 Draw the hyperbola. First draw a rectangle centered at the origin that is 2a = 6 units high and 2b = 4 units wide.

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 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 3 units from the center, so c = 3. The vertices are each 1 units from the center, so a = 1.

GUIDED PRACTICE for Examples 1 and 2 Because c 2 = a 2 + b 2, you have b 2 = c 2 – a 2. Find b 2. b 2 = c 2 – a 2 = 3 2 – 1 2 = 8 Because the transverse axis is horizontal, the standard form of the equation is as follows: x – y28y28 = 1 Substitute 1 for a and 8 for b 2. x 2 – y28y28 = 1 Simplify

GUIDED PRACTICE for Examples 1 and 2 5. Foci: (0, – 10), (0, 10) Vertices: (0, – 6), (0, 6) SOLUTION The foci and vertices lie on the y -axis equidistant from the origin, so the transverse axis is vertical and the center is the origin. The foci are each 10 units from the center, so c = 10. The vertices are each 6 units from the center, so a = 6. Because c 2 = a 2 + b 2, you have b 2 = c 2 – a 2. Find b 2. b 2 = c 2 – a 2 = 10 2 – 6 2 = 64

GUIDED PRACTICE for Examples 1 and 2 Because the transverse axis is horizontal, the standard form of the equation is as follows: y – x 2 64 = 1 Substitute 6 for a and 64 for b 2. Simplify = 1 y 2 36 – x 2 64