Problem 21 - Technical Mathematics with Calculus

Question

Write the equation of each hyperbola in standard form. $$\frac{x^{2}}{25}-\frac{y^{2}}{9}=1$$

Step-by-Step Solution

Answer

The equation of the hyperbola is already in standard form: \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\).

Step 1: Identify the Standard Form of a Hyperbola

Recognize that the given equation has the form of a hyperbola in standard form, which is \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\) for a hyperbola that opens horizontally, or \(\frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1\) for a hyperbola that opens vertically. In the given equation, it is already in the standard form of a hyperbola that opens horizontally.

Step 2: Determine the Asymptotes

Find the asymptotes of the hyperbola using the equations \(y = \pm\frac{b}{a}x\). Here, \(a^2 = 25\) and \(b^2 = 9\), giving us \(a = 5\) and \(b = 3\). Therefore, the equations of the asymptotes are \(y = \pm\frac{3}{5}x\).

Step 3: Identify the Vertices and Foci

Locate the vertices along the x-axis at a distance of \(a\) from the center (0, 0), giving us vertices at \((-5, 0)\) and \((5, 0)\). To find the foci, we use \(c^2 = a^2 + b^2\). Substituting in the given values, we find \(c^2 = 25 + 9 = 34\), so \(c = \sqrt{34}\). Hence, the foci are located at \((-\sqrt{34}, 0)\) and \((\sqrt{34}, 0)\). However, since this step is about writing the equation in standard form and not about locating vertices or foci, this step is not necessary for this task.

Key Concepts

Understanding Hyperbola Asymptotes

When studying hyperbolas, asymptotes play a key role in determining the shape and orientation of the curve. Asymptotes are lines that the hyperbola approaches but never actually meets, no matter how far the branches of the hyperbola extend.

For the hyperbola given by the equation \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\), it opens horizontally as indicated by the position of the variables. To find the asymptotes, we examine the coefficients of \(x^{2}\) and \(y^{2}\). We identify \(a^2\) as 25 and \(b^2\) as 9, which gives us \(a = 5\) and \(b = 3\). The asymptotes can then be determined using the formula \(y = \pm\frac{b}{a}x\), leading to the lines \(y = \pm\frac{3}{5}x\). These diagonal lines provide a skeleton for the hyperbola, giving insight into the limits of its spread.

Locating Hyperbola Vertices

The vertices of a hyperbola are points where the curve changes direction and are also the closest points of the hyperbola to its center. For the given standard equation of a hyperbola, \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\), we can find the vertices because the hyperbola opens along the x-axis.

The value of \(a\) from the equation represents the distance from the center to each vertex along the major axis. With \(a = 5\), our hyperbola's vertices are located at \( (-a, 0)\) and \( (a, 0)\), or specifically, at \( (-5, 0)\) and \( (5, 0)\). These vertices are pivotal in sketching the hyperbola accurately and are integral to understanding its geometry and equations.

Finding the Hyperbola Foci

The foci of a hyperbola are two points located along the major axis that are fundamental in defining the curve's shape. Each point on a hyperbola is equidistant from a focus and the corresponding directrix.

To discover the foci of the hyperbola \(\frac{x^{2}}{25}-\frac{y^{2}}{9}=1\), we employ the relationship \(c^2 = a^2 + b^2\). Given that \(a^2 = 25\) and \(b^2 = 9\), we find \(c^2 = 34\). Thus, the foci are positioned at \( (\pm\sqrt{34}, 0)\), or approximately \( (\pm5.83, 0)\). The foci, while not visible on the graph of a hyperbola, are essential for constructions and for understanding the hyperbola's eccentricity and unique properties.