Question

find all vertices and foci of the hyperbola given by
\\( \frac { ( y - 3 ) ^ { 2 } } { 9 } - \frac { ( x + 2 ) ^ { 2 } } { 7 } = 1 \\)
enter your answers as ordered pairs separated by a comma.
vertices

foci

Explanation:

Step1: Identify the standard form parameters

The standard form of a hyperbola is \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\). For the given hyperbola \(\frac{(y - 3)^2}{9}-\frac{(x + 2)^2}{7}=1\), we have \(h=-2\), \(k = 3\), \(a^2=9\) (so \(a = 3\)), \(b^2=7\).

Step2: Calculate \(c\)

Use the formula \(c^2=a^2 + b^2\). Substitute \(a^2 = 9\) and \(b^2=7\), then \(c^2=9 + 7=16\), so \(c = 4\).

Step3: Find the vertices

The vertices of a hyperbola \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) are \((h,k\pm a)\). Substitute \(h=-2\), \(k = 3\), \(a = 3\). The vertices are \((-2,3 + 3)=(-2,6)\) and \((-2,3-3)=(-2,0)\).

Step4: Find the foci

The foci of a hyperbola \(\frac{(y - k)^2}{a^2}-\frac{(x - h)^2}{b^2}=1\) are \((h,k\pm c)\). Substitute \(h=-2\), \(k = 3\), \(c = 4\). The foci are \((-2,3 + 4)=(-2,7)\) and \((-2,3-4)=(-2,-1)\).

Answer:

Vertices: \((-2,6),(-2,0)\); Foci: \((-2,7),(-2,-1)\)