Question

find the equation of the hyperbola with the following properties. express your answer in standard form. foci at (7, 1) and (15, 1) vertices at (8, 1) and (14, 1)

Explanation:

Step1: Find the center \((h,k)\)

The center of the hyperbola is the mid - point of 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 \((7,1)\) and \((15,1)\) (or vertices \((8,1)\) and \((14,1)\)), \(h=\frac{7 + 15}{2}=\frac{8+14}{2}=11\), \(k = 1\).

Step2: Find \(a\) (distance from center to vertex)

The distance from the center \((11,1)\) to a vertex. Using the vertex \((8,1)\) (or \((14,1)\)), \(a=\vert11 - 8\vert=\vert14 - 11\vert = 3\).

Step3: Find \(c\) (distance from center to focus)

The distance from the center \((11,1)\) to a focus. Using the focus \((7,1)\) (or \((15,1)\)), \(c=\vert11 - 7\vert=\vert15 - 11\vert=4\).

Step4: Find \(b\) using the relationship \(c^{2}=a^{2}+b^{2}\)

Substitute \(a = 3\) and \(c = 4\) into \(c^{2}=a^{2}+b^{2}\).
\(b^{2}=c^{2}-a^{2}\)
\(b^{2}=4^{2}-3^{2}=16 - 9=7\).

Step5: Write the standard form of the hyperbola

Since the foci and vertices have the same \(y\) - coordinate, the hyperbola has a horizontal transverse axis. The standard form is \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}}=1\).
Substitute \(h = 11\), \(k = 1\), \(a^{2}=9\), \(b^{2}=7\) into the formula: \(\frac{(x - 11)^{2}}{9}-\frac{(y - 1)^{2}}{7}=1\).

Answer:

\(\frac{(x - 11)^{2}}{9}-\frac{(y - 1)^{2}}{7}=1\)