Question

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

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

Explanation:

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

For the hyperbola \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), 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 \(h + a\) and \(h - a\)

\(h+a=2 + 4=6\), \(h - a=2-4=-2\).

Answer:

\((-2,1)\), \((6,1)\)