The vertices for the hyperbola $\frac{(y-2)^2}{16}-\frac{(x+1)^2}{144}=1$ are $(-1,6)$ and $(-1,-2)$.
A. True
B. False
Answer (1)
Identify the center ( h , k ) and the values of a and b from the given hyperbola equation.
Calculate the vertices using the formula ( h , k ± a ) .
Compare the calculated vertices with the given vertices.
Conclude that the statement is True because the calculated vertices match the given vertices: T r u e .
Explanation
Analyze the problem The equation of the hyperbola is given as 16 ( y − 2 ) 2 − 144 ( x + 1 ) 2 = 1 . We need to determine if the given vertices ( − 1 , 6 ) and ( − 1 , − 2 ) are correct for this hyperbola.
Identify the center and a, b values The standard form of a hyperbola with a vertical transverse axis is a 2 ( y − k ) 2 − b 2 ( x − h ) 2 = 1 , where ( h , k ) is the center of the hyperbola. In our equation, we can identify the center as ( h , k ) = ( − 1 , 2 ) . Also, a 2 = 16 and b 2 = 144 , which means a = 16 = 4 and b = 144 = 12 .
Find the vertices The vertices of the hyperbola are located at ( h , k ± a ) . Substituting the values of h , k , and a , we get the vertices as ( − 1 , 2 ± 4 ) .
Calculate the coordinates Now, let's calculate the coordinates of the vertices:
Vertex 1: ( − 1 , 2 + 4 ) = ( − 1 , 6 ) Vertex 2: ( − 1 , 2 − 4 ) = ( − 1 , − 2 )
So, the vertices are ( − 1 , 6 ) and ( − 1 , − 2 ) .
Compare and conclude Comparing the calculated vertices ( − 1 , 6 ) and ( − 1 , − 2 ) with the given vertices ( − 1 , 6 ) and ( − 1 , − 2 ) , we see that they match. Therefore, the statement is true.
Examples
Understanding hyperbolas is crucial in various fields. For instance, in physics, the trajectory of a comet as it approaches and recedes from the sun can be modeled using a hyperbola. Similarly, in engineering, the design of cooling towers often involves hyperbolic shapes to optimize structural integrity and airflow. By studying the properties of hyperbolas, such as their vertices and asymptotes, we can better analyze and predict the behavior of these real-world phenomena.
