91Ó°ÊÓ

In a hyperbola, vertices play a crucial role in determining its shape and position. For the hyperbola described by the equation \(9y^2 - 25x^2 = 225\), the vertices are the points where the hyperbola intersects its transverse axis. To find the vertices, we first need to convert the equation to its standard form: \(\frac{y^2}{25} - \frac{x^2}{9} = 1\). This indicates a vertical hyperbola because the \(y\)-term comes first.

For a vertical hyperbola, the vertices are at the points \((0, \pm a)\), where \(a = \sqrt{25} = 5\). Thus, the vertices for this hyperbola are at \((0, 5)\) and \((0, -5)\). The center of the hyperbola is at the origin \((0, 0)\), and the vertices are 5 units away along the \(y\)-axis.

Asymptotes are lines that the hyperbola approaches as it extends infinitely. They create a sort of "guiding boundary" for the shape of the hyperbola. For the given hyperbola equation in its standard form, \(\frac{y^2}{25} - \frac{x^2}{9} = 1\), the asymptotes are determined by the formula \(y = \pm \frac{a}{b}x\).

Here, \(a = 5\) and \(b = 3\). So the equations of the asymptotes are \(y = \pm \frac{5}{3}x\). This means the asymptotes are straight lines with slopes of \(\frac{5}{3}\) and \(-\frac{5}{3}\) that pass through the origin. These lines help us sketch the graph by indicating how the hyperbola opens and stretches.

The foci of a hyperbola are two distinct points located along the transverse axis. They are used to define the hyperbola's shape mathematically. For our hyperbola, \(\frac{y^2}{25} - \frac{x^2}{9} = 1\), the foci are along the \(y\)-axis, as it is a vertical hyperbola.

To find the foci, we use the relationship \(c = \sqrt{a^2 + b^2}\), where \(a = 5\) and \(b = 3\). Calculating \(c\), we have \(c = \sqrt{5^2 + 3^2} = \sqrt{34}\). Therefore, the foci are at the coordinates \((0, \pm \sqrt{34})\). These points are crucial for applications such as satellite dishes, which often rely on hyperbolic shapes to focus signals.

The standard form of a hyperbola's equation helps us identify its key features easily. A hyperbola can either open horizontally or vertically, and its equation reflects this. Our exercise started with \(9y^2 - 25x^2 = 225\).

To convert this equation into the standard form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) (for a vertical hyperbola), we divide each term by 225 to yield \(\frac{y^2}{25} - \frac{x^2}{9} = 1\). This tells us:

• The hyperbola is vertical because the \(y\)-term comes first.
• \(a^2 = 25\) and \(b^2 = 9\), from which we derive \(a = 5\) and \(b = 3\).

Understanding the standard form allows us to quickly calculate vertices, foci, and asymptotes, and visualize the hyperbola's graph effectively.