Question

1. use the equation $\frac{(x - 2)^{2}}{4^{2}}-\frac{(y - 1)^{2}}{2^{2}} = 1$ to find the vertices.

vertices= $(h + a,k)$ and $(h - a,k)$
$(2,1)$
$(1,6)$
$(-2,1)$
$(6,1)$
$(1,-2)$

Explanation:

Step1: Identify \(h\), \(k\), and \(a\)

For the hyperbola equation \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), comparing with \(\frac{(x - 2)^{2}}{4^{2}}-\frac{(y - 1)^{2}}{2^{2}}=1\), we have \(h = 2\), \(k = 1\), \(a = 4\).

Step2: Calculate the vertices

Using the formula for vertices \((h + a,k)\) and \((h - a,k)\).
For \((h + a,k)\): \(h+a=2 + 4=6\), so the point is \((6,1)\).
For \((h - a,k)\): \(h - a=2-4=-2\), but this is not among the options.

Answer:

\((6,1)\)