Question

f each. then sketch the graph.
$(0,0)$ $\frac{y^2}{4} - \frac{x^2}{4} = 1$
standard form equation of each hyperbola.

1. vertices: $(9, 0)$, $(-9, 0)$

foci: $(\sqrt{130}, 0)$, $(-\sqrt{130}, 0)$

Explanation:

Problem 22: Find the standard form equation of the hyperbola with vertices \((9, 0)\), \((-9, 0)\) and foci \((\sqrt{130}, 0)\), \((-\sqrt{130}, 0)\)

Step 1: Identify the orientation and \(a\)

The vertices are on the \(x\)-axis (\(y = 0\)), so the hyperbola opens horizontally. The standard form is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). The distance from the center (origin, since vertices are \((\pm 9, 0)\)) to a vertex is \(a\), so \(a = 9\), thus \(a^2 = 81\).

Step 2: Find \(c\) and then \(b^2\)

The distance from the center to a focus is \(c\). Here, \(c = \sqrt{130}\), so \(c^2 = 130\). For hyperbolas, \(c^2 = a^2 + b^2\). Solve for \(b^2\):
\(b^2 = c^2 - a^2 = 130 - 81 = 49\).

Step 3: Write the standard equation

Substitute \(a^2 = 81\) and \(b^2 = 49\) into the horizontal hyperbola standard form:
\(\frac{x^2}{81} - \frac{y^2}{49} = 1\).

Answer:

\(\boldsymbol{\frac{x^2}{81} - \frac{y^2}{49} = 1}\)