Problem 73 - Precalculus: Functions and Graphs

Question

The graphs of the equations $$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \quad \text { and } \quad \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1$$ are called conjugate hyperbolas. Sketch the graphs of both equations on the same coordinate plane, with \(a=5\) and \(b=3,\) and describe the relationship between the two graphs.

Step-by-Step Solution

Answer

The two hyperbolas open perpendicularly to each other and share asymptotes.

Step 1: Understand the Equations

The first equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) is the equation of a hyperbola centered at the origin with its transverse axis along the x-axis. The second equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\) represents a hyperbola with a transverse axis along the y-axis when \(a\) and \(b\) are interchanged.

Step 2: Identify Parameters

Set \(a=5\) and \(b=3\). Substitute these into the equations: \(\frac{x^2}{25} - \frac{y^2}{9} = 1\) and \(\frac{x^2}{25} - \frac{y^2}{9} = -1\). These define two hyperbolas, one horizontal and one vertical.

Step 3: Sketch the First Hyperbola

For the equation \(\frac{x^2}{25} - \frac{y^2}{9} = 1\), the vertices are at \((\pm a, 0) = (\pm 5, 0)\) and the asymptotes are the lines \(y = \pm \frac{3}{5}x\). Draw the hyperbola opening horizontally.

Step 4: Sketch the Second Hyperbola

For the equation \(\frac{x^2}{25} - \frac{y^2}{9} = -1\), rearrange it to \(-\frac{x^2}{25} + \frac{y^2}{9} = 1\), indicating the vertices are at \((0, \pm b) = (0, \pm 3)\) and the asymptotes remain \(y = \pm \frac{3}{5}x\). Draw the hyperbola opening vertically.

Step 5: Analyze the Relationship

Notice that the two hyperbolas share the same asymptotes and are reflections of one another across the line \(y=x\) or \(y=-x\). They are orthogonal to each other because one opens horizontally and the other vertically, and they share the same center at the origin.

Key Concepts

Hyperbola Equation

The equations given are standard hyperbola equations:

• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\)

These are known as conjugate hyperbolas. Each equation describes a hyperbola that is centered at the origin. For the first equation, the transverse axis lies along the x-axis, making it a horizontal hyperbola. Meanwhile, the second equation, by rearranging or flipping the signs, suggests a focus along the y-axis, marking it as a vertical hyperbola. Conjugate hyperbolas are essentially reflections of each other, showcasing their unique relationship.

Transverse Axis

In hyperbolas, the transverse axis plays a vital role in defining their shape and orientation.When the equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the transverse axis is horizontal, passing through the vertices on the x-axis. This is because the positive term is associated with the \(x^2\), indicating width along the x-axis.

• Transverse axis is along the x-axis: Horizontal hyperbola

When the equation changes to \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\), switching the roles of x and y, the transverse axis is now along the y-axis, making it a vertical hyperbola.Understanding the axis of symmetry helps graph and predict the hyperbola's orientation effectively.

Vertices and Asymptotes

Vertices are where the hyperbola intersects its transverse axis and they define the 'width' of the hyperbola. For the horizontal hyperbola \(\frac{x^2}{25} - \frac{y^2}{9} = 1\), its vertices are at

• \((\pm 5, 0)\)

For the vertical hyperbola rearranged as \(-\frac{x^2}{25} + \frac{y^2}{9} = 1\), vertices shift to

• \((0, \pm 3)\)

Asymptotes are lines that the hyperbola approaches but never touches. They cross each other at the center of the hyperbola, and are shared between both hyperbolas. With slopes for both equations given by \(y = \pm \frac{3}{5}x\), they guide the hyperbola's 'opening' direction.

Graph Sketching

Graphing a hyperbola involves identifying its transverse axis, vertices, and asymptotes, following these patterns to plot:

• Start with the center at the origin \((0,0)\)
• Plot vertices based on the equation (e.g., \(\pm 5\) on the \(x\)-axis for a horizontal hyperbola)
• Draw asymptotes through the origin, using calculated slopes

Once the basic framework is in place, sketch the two branches of the hyperbola approaching these asymptotes, ensuring you are drawing them relative to the correct axis of symmetry (horizontal or vertical). Use light pencil strokes initially to refine the shape gradually.

Reflection Across Axes

Conjugate hyperbolas have a fascinating relationship, often described in terms of symmetry. The horizontal hyperbola \(\frac{x^2}{25} - \frac{y^2}{9} = 1\) can be considered a reflection across either the line \(y = x\) or \(y = -x\) to create the vertical hyperbola, and vice-versa. This reflection explains their shared characteristics:

• They have the same center at the origin
• Share identical asymptotes crossing at the origin

Reflective symmetry gives conjugate hyperbolas their unique appearance, linking their open branches orthogonally (at right angles) to one another, illustrating elegance in geometry.