Find the vertex form of the hyperbola given by the equation \(16x^2 - 9y^2 = 144\).
Answer
Step 3: The form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) indicates that the hyperbola opens to the left and right. The distance from the center to each vertex is 'a' units. Here, 'a' is the square root of the denominator under 'x', so 'a' is \(\sqrt{9} = 3\). Therefore, the vertices of the hyperbola are at (±3, 0).
Step 1 ：Step 1: Divide all terms by 144 to get the equation in the standard form. The standard form of the equation of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for hyperbolas that open left and right, and \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for hyperbolas that open up and down. So, we have \(\frac{x^2}{9} - \frac{y^2}{16} = 1\).
Step 2 ：Step 2: From the standard form, we can see that the center of the hyperbola is at the origin (0, 0).
Step 3 ：Step 3: The form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) indicates that the hyperbola opens to the left and right. The distance from the center to each vertex is 'a' units. Here, 'a' is the square root of the denominator under 'x', so 'a' is \(\sqrt{9} = 3\). Therefore, the vertices of the hyperbola are at (±3, 0).
