Question

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

$(1,-3)$
$(-3,3)$
$(3,1)$
$(3,-3)$
$(-1,-3)$

Explanation:

Step1: Recall the standard form of a 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 \((h,k)\) and vertices \((h\pm a,k)\).
For the given hyperbola \(\frac{(x - 1)^{2}}{4}-\frac{(y + 3)^{2}}{16}=1\), we have \(h = 1\), \(k=-3\), and \(a^{2}=4\), so \(a = 2\).

Step2: Calculate the vertices

Using the formula for vertices \((h\pm a,k)\), substitute \(h = 1\), \(a = 2\), and \(k=-3\).
For \(x\) - coordinate: \(x=1\pm2\).
When \(x=1 + 2=3\), \(y=-3\); when \(x=1-2=-1\), \(y=-3\).

Answer:

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