Identify the center ( h , k ) , a 2 , and b 2 from the hyperbola equation.
Calculate c using c 2 = a 2 + b 2 , which gives c = 10 .
Determine the foci using ( h ± c , k ) , resulting in foci at ( 12 , − 1 ) and ( − 8 , − 1 ) .
Compare the calculated foci with the given foci and conclude that the statement is false. F a l se
Explanation
Analyze the problem We are given the equation of a hyperbola as 36 ( x − 2 ) 2 − 64 ( y + 1 ) 2 = 1 . We need to determine if the given foci ( 2 , − 1 + 4 7 ) and ( 2 , − 1 − 4 7 ) are correct for this hyperbola.
Identify the parameters The standard form of a hyperbola centered at ( h , k ) with a horizontal transverse axis is a 2 ( x − h ) 2 − b 2 ( y − k ) 2 = 1 . The foci are located at ( h ± c , k ) , where c 2 = a 2 + b 2 . In our case, the hyperbola is centered at ( 2 , − 1 ) , a 2 = 36 , and b 2 = 64 .
Calculate c First, we need to find the value of c . We know that c 2 = a 2 + b 2 , so c 2 = 36 + 64 = 100 . Taking the square root of both sides, we get c = 100 = 10 .
Determine the foci Since the hyperbola has a horizontal transverse axis, the foci are located at ( h ± c , k ) . Substituting the values h = 2 , k = − 1 , and c = 10 , we find the foci to be ( 2 ± 10 , − 1 ) , which are ( 2 + 10 , − 1 ) = ( 12 , − 1 ) and ( 2 − 10 , − 1 ) = ( − 8 , − 1 ) .
Compare and conclude The calculated foci are ( 12 , − 1 ) and ( − 8 , − 1 ) . The given foci are ( 2 , − 1 + 4 7 ) and ( 2 , − 1 − 4 7 ) . These are not the same. Therefore, the statement is false.
Examples
Understanding hyperbolas is crucial in various fields, such as astronomy, where they describe the paths of comets as they approach and recede from the sun. Also, the properties of hyperbolas are used in the design of lenses and mirrors in telescopes and other optical instruments. By knowing the equation of a hyperbola, we can determine its key features, like the foci, which helps in accurately predicting the trajectory of celestial objects or optimizing the design of optical systems.
The calculated foci of the hyperbola 36 ( x − 2 ) 2 − 64 ( y + 1 ) 2 = 1 are ( 12 , − 1 ) and ( − 8 , − 1 ) . These do not match the given foci of ( 2 , − 1 + 4 7 ) and ( 2 , − 1 − 4 7 ) . Therefore, the statement is False.
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