Question

1. find the center of the hyperbola with equation \\( \frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \\).

\\( ( 3, - 1 ) \\)
\\( ( - 1,3 ) \\)
\\( ( - 3,1 ) \\)
\\( ( 1, - 3 ) \\)

1. find the vertices of the hyperbola \\( \frac { ( x - 1 ) ^ { 2 } } { 4 } - \frac { ( y + 3 ) ^ { 2 } } { 16 } = 1 \\).

Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\) (opens left - right) has center \((h,k)\).

Step2: Identify \(h\) and \(k\)

For the given hyperbola \(\frac{(x - 1)^{2}}{4}-\frac{(y+3)^{2}}{16}=1\), we have \(x - 1=x - h\) and \(y + 3=y - k\).
So, \(h = 1\) and \(k=-3\).

Answer:

\((1,-3)\)