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Solve Applied Problems Involving Hyperbolas
-A Satellite Following the Hyperbolic (0,6)( 0,6 )

Question 83

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Solve Applied Problems Involving Hyperbolas
-A satellite following the hyperbolic path shown in the picture turns rapidly at (0,6) ( 0,6 ) and then moves closer and closer to the line y=92xy = \frac { 9 } { 2 } x as it gets farther from the tracking station at the origin. Find the equation that describes the path of the satellite if the center of the hyperbola is at (0,0) ( 0,0 ) .
 Solve Applied Problems Involving Hyperbolas -A satellite following the hyperbolic path shown in the picture turns rapidly at  ( 0,6 )   and then moves closer and closer to the line  y = \frac { 9 } { 2 } x  as it gets farther from the tracking station at the origin. Find the equation that describes the path of the satellite if the center of the hyperbola is at  ( 0,0 )  .   A)   \frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { \frac { 16 } { 9 } } = 1  B)   \frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { \left( \frac { 54 } { 4 } \right)  ^ { 2 } } = 1  C)   \frac { y ^ { 2 } } { \frac { 16 } { 9 } } - \frac { x ^ { 2 } } { 36 } = 1  D)   \frac { x ^ { 2 } } { \left( \frac { 54 } { 4 } \right)  ^ { 2 } } - \frac { y ^ { 2 } } { 36 } = 1


A) y236x2169=1\frac { y ^ { 2 } } { 36 } - \frac { x ^ { 2 } } { \frac { 16 } { 9 } } = 1
B) x236y2(544) 2=1\frac { x ^ { 2 } } { 36 } - \frac { y ^ { 2 } } { \left( \frac { 54 } { 4 } \right) ^ { 2 } } = 1
C) y2169x236=1\frac { y ^ { 2 } } { \frac { 16 } { 9 } } - \frac { x ^ { 2 } } { 36 } = 1
D) x2(544) 2y236=1\frac { x ^ { 2 } } { \left( \frac { 54 } { 4 } \right) ^ { 2 } } - \frac { y ^ { 2 } } { 36 } = 1

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