Question

find the vertices of the hyperbola. enter the smallest x coordinate first.

$$ \frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 40 } = 1 $$

$$ ( square, square ) \text { and } ( square, square ) $$

Explanation:

Step1: Recall the standard form of hyperbola

The standard form of a hyperbola centered at the origin \((0,0)\) with a horizontal transverse axis is \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1\). The vertices are \((\pm a,0)\).

Step2: Identify \(a^{2}\) from the given equation

Given \(\frac{x^{2}}{9}-\frac{y^{2}}{40}=1\), we have \(a^{2}=9\). Then \(a = 3\) (since \(a>0\)).

Answer:

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