Problem 54 - Algebra 2

Question

Write the equation of each hyperbola in standard form. Sketch the graph. $$ x^{2}-25 y^{2}=25 $$

Step-by-Step Solution

Answer

The standard form of the equation is \(\frac{x^{2}}{5^{2}} - \frac{y^{2}}{1^{2}} = 1\). The vertices of the hyperbola are at (-5,0) and (5,0), and the foci are at (-sqrt(26),0) and (sqrt(26),0). The asymptotes are \(y= \pm \frac{1}{5}x\). The hyperbola opens left and right.

Step 1: Rewrite the equation in standard form

The equation is given as \(x^{2}-25 y^{2}=25\). To rewrite this in standard form, we need to divide through by 25. The result is \[ \frac{x^{2}}{25} - y^{2} = 1 \], which can further be rewritten as \[\frac{x^{2}}{5^{2}} - \frac{y^{2}}{1^{2}} = 1\]. So the standard form of the equation is \[\frac{x^{2}}{5^{2}} - \frac{y^{2}}{1^{2}} = 1\].

Step 2: Calculate the values of a, b and c.

From the standard form of the equation, we can see that \(a=5\) and \(b=1\). The distance c from the center to each focus can be calculated using the formula \(c = \sqrt{a^{2}+b^{2}}\), which yields \(c = \sqrt{5^{2}+1^{2}} = \sqrt{26}\).

Step 3: Sketch the graph

The hyperbola opens left and right because the x-term is positive. Its center is at the origin. The vertices are at (-5,0) and (5,0), because the distance from the center to each vertex is a, and the foci are at (-sqrt(26),0) and (sqrt(26),0). We then draw the asymptotes, which are the lines \(y= \pm \frac{b}{a}x = \pm \frac{1}{5}x \), and sketch the hyperbola approaching these asymptotes.

Key Concepts

Standard form equation

The standard form equation of a hyperbola is crucial to understanding its characteristics and graphing it accurately. A hyperbola takes the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) if it opens left and right, or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) if it opens up and down. The equation involves two key variables, \(a\) and \(b\), which determine the shape and orientation of the hyperbola.

In this particular exercise, we begin with the equation \(x^2 - 25y^2 = 25\). To convert this into the standard form, we divide every term by 25, yielding \(\frac{x^2}{25} - \frac{y^2}{1} = 1\). It simplifies further to \(\frac{x^2}{5^2} - \frac{y^2}{1^2} = 1\).

This final expression clearly indicates that \(a^2 = 25\) (thus \(a = 5\)) and \(b^2 = 1\) (thus \(b = 1\)). These values are used to identify various properties of the hyperbola, such as the vertices and asymptotes.

Asymptotes

Asymptotes are lines that a hyperbola approaches but never actually touches. For hyperbolas, these lines are defined by the slopes \(\pm \frac{b}{a}\) and they guide the shape of the hyperbola as it extends toward infinity.

In the given exercise, we found that \(a = 5\) and \(b = 1\). This means the slopes of our asymptotes are \(\pm \frac{b}{a} = \pm \frac{1}{5}\). Therefore, the equations of the asymptotes are
\[y = \frac{1}{5}x \quad \text{and} \quad y = -\frac{1}{5}x\].
These asymptotes intersect at the center of the hyperbola, which in this case is the origin, (0,0).

When sketching, these lines serve as guidelines towards which the branches of the hyperbola curve but never meet. This guidance is vital for understanding the broad characteristics of the hyperbola's graph.

Vertices of hyperbola

Vertices are the points on the hyperbola that are closest to or furthest from the center, depending on the orientation. They provide a way to measure the "spread" of the hyperbola.

For the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the vertices lie along the x-axis when the hyperbola opens left and right.
From the calculations, we know \(a = 5\), which tells us the distance from the center to each vertex is 5 units. Thus, the vertices of the hyperbola are located at:

• (-5, 0)
• (5, 0)

These points are directly on the hyperbola's transverse axis, a central feature of hyperbolas, helping to define their shape and orientation.

In graphing, locating the vertices informs us of the hyperbola's direction and width across the transverse axis.

Foci of hyperbola

The foci of a hyperbola are points from which distances are measured to determine the shape of the hyperbola. They are essential in defining the hyperbola's eccentricity and overall shape.

Given the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the foci of the hyperbola can be found along the x-axis as it opens left and right. From the calculations, the distance \(c\) to each focus is obtained by \(c = \sqrt{a^2 + b^2}\).
This leads to
\[c = \sqrt{5^2 + 1^2} = \sqrt{26}\].
Hence, the coordinates of the foci are:

• (-\(\sqrt{26}\), 0)
• (\(\sqrt{26}\), 0)

These points help draw the imaginary arms of the hyperbola, defining how it spreads and bends around the center.

Locating the foci provides deeper insight into the geometry and balance of the hyperbola's graph.