Problem 3 Modeling with a Hyperbola

Communications The graph shows a 2-dimensional view of a satellite dish. The focus is located at F 1 f sub 1  but the receiving device is located on the bottom of the dish at the point F 2 . f sub 2 , .  The rays are reflected by the first reflector (the black curve), toward F 1 f sub 1  and then reflected by the second reflector (the blue curve) toward F 2 . f sub 2 , .

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1. What kind of curve is the second reflector? How can you tell?

   Think

   What information does the diagram give up?

   It helps you see the relative positions of the reflectors and foci.

   The second reflector is a hyperbola because it reflects rays aimed at its internal focus toward its external focus.

2. The vertex of the second reflector is 3 in. from F 1 f sub 1  and 21 in. from F 2 . f sub 2 , .  What is an equation for the second reflector? Assume the conic is horizontal and centered at the origin.

   Step 1	Determine the standard-form equation of the conic. The conic is a horizontal hyperbola centered at the origin. x 2 a 2 − y 2 b 2 = 1 fraction x squared , over eh squared end fraction . minus . fraction y squared , over b squared end fraction . equals 1
   Step 2	Find c. The distance between the foci is 3 + 21 = 24 in. Since c is half this distance, c = 12 . c equals 12 .
   Step 3	Find a. The distance from the internal focus to the vertex of the reflector is 3 in. So, c − a = 12 − a = 3 c minus eh equals 12 minus eh equals 3  and a = 9 . eh equals 9 .
   Step 4	Use c and a to find b 2 . b squared , . b 2 = c 2 − a 2     = 12 2 − 9 2 Substitute and simplify.   = 63 Simplify. table with 3 rows and 3 columns , row1 column 1 , b squared , column 2 equals , c squared , minus , eh squared , column 3 , row2 column 1 , , column 2 equals , 12 squared , minus , 9 squared , column 3 cap substituteandsimplify. , row3 column 1 , , column 2 equals 63 , column 3 cap simplify. , end table
   Step 5	Use a and b 2 b squared  to write the equation. An equation is x 2 81 − y 2 63 = 1 . fraction x squared , over 81 end fraction , minus , fraction y squared , over 63 end fraction , equals 1 .

✓ Got It?

3. Suppose the vertex of the second reflector in Problem 3 were 4 in. from F 1 f sub 1  and 18 in. from F 2 . f sub 2 , .  What is the equation for the second reflector? Assume the conic is horizontal and centered at the origin.