Problem 65 - A Graphical Approach to College Algebra

Question

Find an equation for each hyperbola. Asymptotes \(y=\pm \frac{3}{5} x ; y\) -intercepts \((0, \pm 3)\)

Step-by-Step Solution

Answer

The equation is \(\frac{x^2}{25} - \frac{y^2}{9} = 1\).

Step 1: Identify the Hyperbola Orientation

Since the asymptotes are given in the form \(y = \pm \frac{3}{5}x\), the hyperbola opens horizontally. This means that the general equation of the hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Step 2: Use the Asymptote Slopes

The slopes of the asymptotes are given by \(\pm \frac{b}{a}\). Thus, \(\frac{b}{a} = \frac{3}{5}\). From the asymptotes, we can deduce that \(b = \frac{3}{5}a\).

Step 3: Substitute the y-intercepts

The y-intercepts are \((0, \pm 3)\). Substituting \((0,3)\) into the hyperbola equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we get \(\frac{0^2}{a^2} - \frac{3^2}{b^2} = 1\), resulting in \(- \frac{9}{b^2} = 1\).

Step 4: Solve for b

From \(- \frac{9}{b^2} = 1\), solve for \(b^2\). We get \(-9 = b^2\), hence \(b^2 = 9\) as the sign following division adjustment. Thus, \(b = 3\).

Step 5: Use the Relation to Find a

Since we have \(b = 3\) and \(b = \frac{3}{5}a\), we substitute to find \(a\). Therefore, \(3 = \frac{3}{5}a\). Solving for \(a\), we get \(a = 5\).

Step 6: Final Equation

Now that we have \(a = 5\) and \(b = 3\), substitute these into the general equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). The equation becomes \(\frac{x^2}{25} - \frac{y^2}{9} = 1\).

Key Concepts

Asymptotes

Asymptotes play a crucial role in defining the behavior of a hyperbola. They are imaginary diagonal lines that the hyperbola approaches but never crosses. In the given hyperbola, the asymptotes are represented by the equations \( y = \pm \frac{3}{5}x \). From these lines, we understand how the arms of the hyperbola are oriented. The presence of \( \pm \) indicates two lines that form a symmetric 'X' pattern, with one line having a positive slope (\( \frac{3}{5} \)) and the other a negative slope (\( -\frac{3}{5} \)). This pairing of slopes helps in determining the direction in which the hyperbola will open.

Hyperbola Orientation

The orientation of a hyperbola refers to the direction in which it opens. When the slopes of the asymptotes, given as \( y = \pm \frac{3}{5}x \), are derived from a set of hyperbolic equations, it indicates how the arms of the hyperbola diverge. Here, since the asymptotes are expressed with the term \( x \), and the expression involves a horizontal ratio \( \frac{3}{5} \), the hyperbola opens horizontally. This is reflective in the hyperbola equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where the \( x^2 \) fraction is positive, supporting a horizontal orientation. With a horizontal hyperbola, the arms extend left and right along the x-axis.

Y-intercepts

Y-intercepts are the points where a graph crosses the y-axis, providing valuable information about the hyperbola's width and positioning. For this particular hyperbola, the y-intercepts are given as \( (0, \pm 3) \). This means that when \( x = 0 \), the hyperbola crosses the y-axis at \( y = 3 \) and \( y = -3 \). These intercepts help us confirm the hyperbola's dimensions and validate our calculations of \( b \), where substituting \( y = 3 \) confirms \( \frac{x^2}{a^2} - \frac{3^2}{b^2} = 1 \). It provides a checkpoint for confirming the values of \( a \) and \( b \) derived during the solution process.

Equation Solving

Equation solving for hyperbolas involves using the given characteristics—such as asymptotes and y-intercepts—to find the equation's unknowns. From the asymptote slopes \( \frac{b}{a} = \frac{3}{5} \) and one y-intercept \( (0, 3) \), we can set up equations to solve for \( a \) and \( b \). First, calculate \( b^2 \) using the intercept \( -\frac{9}{b^2} = 1 \), solving to find \( b = 3 \). Then use \( b = \frac{3}{5}a \) to find \( a = 5 \). Finally, substitute these values into the hyperbola's standard form to get \( \frac{x^2}{25} - \frac{y^2}{9} = 1 \). This systematic approach allows us to accurately develop the equation from given data.