Question

find the vertices of the hyperbola. enter the smallest x coordinate first.
$$\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 40 } = 1$$
( ? , ) and ( , )

Explanation:

Step1: Identify the standard form of hyperbola

The standard form of a hyperbola centered at the origin \((h = 0,k = 0)\) with a horizontal transverse axis is \(\frac{(x - h)^2}{a^2}-\frac{(y - k)^2}{b^2}=1\). For the given equation \(\frac{x^{2}}{9}-\frac{y^{2}}{40}=1\), we have \(h = 0,k = 0,a^{2}=9,b^{2}=40\).

Step2: Find the value of \(a\)

Since \(a^{2}=9\), then \(a=\sqrt{9}=3\).

Step3: Determine the vertices

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

Answer:

\((-3,0)\) and \((3,0)\)