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EXAMPLE 1

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

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EXAMPLE 1

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  1. 100 y2 25y2 4x2 100 4 100 100 = – y225 – = 1 EXAMPLE 1 Graph an equation of a hyperbola Graph 25y2 – 4x2 = 100. Identify the vertices, foci, and asymptotes of the hyperbola. SOLUTION STEP 1 Rewrite the equation in standard form. 25y2 – 4x2 = 100 Write original equation. Divide each side by 100. Simplify.

  2. 29. 29. soc = The foci are at( 0, + ) (0, + 5.4). 25 ab + + The asymptotes are y = x x y = or EXAMPLE 1 Graph an equation of a hyperbola STEP 2 Identify the vertices, foci, and asymptotes. Note that a2 = 4 and b2 = 25, so a = 2and b = 5. The y2 - term is positive, so the transverse axis is vertical and the vertices are at (0, +2). Find the foci. c2= a2– b2= 22– 52= 29.

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

  4. 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, soa = 3.

  5. = 1 x2 x2 9 32 = 1 y2 7 y2 7 – – EXAMPLE 2 Write an equation of a hyperbola Becausec2 = a2 + b2, you haveb2 = c2 – a2. Findb2. b2= c2–a2= 42–32= 7 Because the transverse axis is horizontal, the standard form of the equation is as follows: Substitute 3 for aand 7 for b2. Simplify

  6. You can take panoramic photographs using a hyperbolic mirror. Light rays heading toward the focus behind the mirror are reflected to a camera positioned at the other focus as shown. After a photograph is taken, computers can “unwrap” the distorted image into a 360° view. EXAMPLE 3 Solve a multi-step problem Photography

  7. – – y2 7.90 x25.50 y22.812 x2 5.50 b2= c2–a2 = 3.662–2.8125.50 = 1 = 1 EXAMPLE 3 Solve a multi-step problem • Write an equation for the cross section of the mirror. • The mirror is 6 centimeters wide. How tall is it? SOLUTION STEP 1 From the diagram, a= 2.81andc= 3.66. To write an equation, find b2. Because the transverse axis is vertical, the standard form of the equation for the cross section of the mirror is as follows: or

  8. y2 7.90 32 5.50 – = 1 y2 20.83 y – 4.56 ANSWER So, the mirror has a height of– 2.81 – (– 4.56) = 1.75centimeters. EXAMPLE 3 Solve a multi-step problem STEP 2 Find the y-coordinate at the mirror’s bottom edge. Because the mirror is 6 centimeters wide, substitute x = 3 into the equation and solve. Substitute 3 for x. Solve fory2. Solve fory.

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