Solve the problem.
- Radio signals emitted from points (10, 0) and (-10, 0) indicate that a plane is 8 miles closer to (10,0) than to (-10, 0). Find an equation of the hyperbola that passes through the plane's location with foci (10, 0) and (-10, 0). All units are in miles.
Radio signals emitted from points (0, 8) and (0, -8) indicate that a plane is 4 miles farther from (0, 8) than (0, -8). Find an equation of the hyperbola that passes through the plane's location with foci (0,8) and (0, -8).
From the figure, the plane is located in the fourth quadrant of the coordinate system. Solve the system of equations defining the two hyperbolas for the point of intersection in the fourth quadrant. This is the location o the plane.
Round the coordinates to the nearest tenth of a mile.
$Solve the problem. - Radio signals emitted from points (10, 0) and (-10, 0) indicate that a plane is 8 miles closer to (10,0) than to (-10, 0). Find an equation of the hyperbola that passes through the plane's location with foci (10, 0) and (-10, 0). All units are in miles. Radio signals emitted from points (0, 8) and (0, -8) indicate that a plane is 4 miles farther from (0, 8) than (0, -8). Find an equation of the hyperbola that passes through the plane's location with foci (0,8) and (0, -8). From the figure, the plane is located in the fourth quadrant of the coordinate system. Solve the system of equations defining the two hyperbolas for the point of intersection in the fourth quadrant. This is the location o the plane. Round the coordinates to the nearest tenth of a mile. A) a. \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1 b. \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 60 } = 1 c. ( 4.1,2.3 ) B) \mathrm { a } \cdot \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1 b. \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 60 } = 1 c. ( 4.1 , - 2.3 ) C) a. \frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 84 } = 1 b. \frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 60 } = 1 c. ( - 2.3,4.1 ) D) a. \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1 b. \frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 60 } = 1 c. ( 4.1 , - 2.3 )$

A) a.

\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1

\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 60 } = 1

( 4.1,2.3 )

\mathrm { a } \cdot \frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1

\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 60 } = 1

( 4.1 , - 2.3 )

C) a.

\frac { y ^ { 2 } } { 16 } - \frac { x ^ { 2 } } { 84 } = 1

\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 60 } = 1

( - 2.3,4.1 )

D) a.

\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 84 } = 1

\frac { y ^ { 2 } } { 4 } - \frac { x ^ { 2 } } { 60 } = 1

( 4.1 , - 2.3 )

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