Question

find the equation of the hyperbola with the following properties. express your answer in standard form.
foci at (-3,-3) and (-3,-9)
vertices at (-3,-5) and (-3,-7)

Explanation:

Step1: Find the center of the hyperbola

The center \((h,k)\) of a hyperbola is the mid - point between the foci (or vertices).
For two points \((x_1,y_1)\) and \((x_2,y_2)\), the mid - point formula is \((\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2})\).
Using the foci \((-3,-3)\) and \((-3,-9)\):
\(h=\frac{-3+( - 3)}{2}=-3\), \(k=\frac{-3+( - 9)}{2}=\frac{-12}{2}=-6\)
So, the center \((h,k)=(-3,-6)\)

Step2: Determine the values of \(a\) and \(c\)

The distance between the center \((h,k)\) and a vertex is \(a\).
Using the vertex \((-3,-5)\) (or \((-3,-7)\)) and the center \((-3,-6)\):
\(a=\vert-5-( - 6)\vert = 1\) (or \(a=\vert-7-( - 6)\vert = 1\))
The distance between the center \((h,k)\) and a focus is \(c\).
Using the focus \((-3,-3)\) (or \((-3,-9)\)) and the center \((-3,-6)\):
\(c=\vert-3-( - 6)\vert = 3\) (or \(c=\vert-9-( - 6)\vert = 3\))

Step3: Find the value of \(b\)

For a hyperbola, the relationship \(c^{2}=a^{2}+b^{2}\) holds.
Substitute \(a = 1\) and \(c = 3\) into the formula:
\(3^{2}=1^{2}+b^{2}\)
\(b^{2}=c^{2}-a^{2}=9 - 1=8\)

Step4: Write the equation of the hyperbola

Since the foci and vertices have the same \(x\) - coordinate, the hyperbola has a vertical transverse axis.
The standard form of a hyperbola with a vertical transverse axis is \(\frac{(y - k)^{2}}{a^{2}}-\frac{(x - h)^{2}}{b^{2}}=1\)
Substitute \(h=-3,k=-6,a = 1,b^{2}=8\) into the formula:
\(\frac{(y+6)^{2}}{1^{2}}-\frac{(x + 3)^{2}}{8}=1\)

Answer:

\(\frac{(y + 6)^{2}}{1}-\frac{(x + 3)^{2}}{8}=1\)