Problem 34 - Thomas Calculus

Question

Exercises \(27-34\) give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. $$ 64 x^{2}-36 y^{2}=2304 $$

Step-by-Step Solution

Answer

The hyperbola in standard form is \(\frac{x^2}{36} - \frac{y^2}{64} = 1\). Asymptotes are \(y = \pm \frac{4}{3}x\).

Step 1: Identify the given equation and recognize it's for a hyperbola

The equation given is \(64x^2 - 36y^2 = 2304\). This is a quadratic equation in \(x\) and \(y\). The terms \(x^2\) and \(-y^2\) indicate that this is an equation for a hyperbola.

Step 2: Transform the equation to have a right-hand side of 1

Divide the whole equation by 2304 to have the equation in a form where the right-hand side equals 1:\[ \frac{64x^2}{2304} - \frac{36y^2}{2304} = 1. \]

Step 3: Simplify the equation to standard form

Simplify the fractions:\[ \frac{x^2}{36} - \frac{y^2}{64} = 1. \]This is the standard form of the hyperbola, which is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where \(a^2 = 36\) and \(b^2 = 64\). So, \(a = 6\) and \(b = 8\).

Step 4: Find the equations for the asymptotes

For a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations for the asymptotes are \(y = \pm \frac{b}{a}x\).Substitute \(a = 6\) and \(b = 8\):\[ y = \pm \frac{8}{6}x, \]which simplifies to:\[ y = \pm \frac{4}{3}x. \]

Step 5: Sketch the hyperbola, its asymptotes, and foci

- The asymptotes, \(y = \pm \frac{4}{3}x\), should be drawn as diagonal lines through the origin.- The vertices of the hyperbola, for \(\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1\), are at \((\pm 6, 0)\).- Concepts of hyperbola will place foci using \(c^2 = a^2 + b^2\), giving \(c = \sqrt{36 + 64} = \sqrt{100} = 10\), so foci are at \((\pm 10, 0)\).- Sketch the hyperbola opening left and right from the center at the origin, intersecting the x-axis at vertices, and approaching the asymptotes without touching them.

Key Concepts

Asymptotes

In the realm of hyperbolas, asymptotes are invisible lines that the curve approaches but never touches. Visualizing asymptotes helps in sketching a hyperbola correctly. They guide the shape of the hyperbolic curve and give us a general idea of its "spread."

For a hyperbola whose standard form is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are found using the formula \(y = \pm \frac{b}{a}x\). This represents two diagonal lines crossing through the origin.

• In our example, \(a = 6\) and \(b = 8\).
• Thus, the asymptotes are \(y = \pm \frac{8}{6}x\), which simplify to \(y = \pm \frac{4}{3}x\).

This means that the hyperbola's branches will approach these lines as \(x\) goes to infinity in either direction. Asymptotes are key to determining the direction and steepness of a hyperbola's branches.

Standard Form of Hyperbola

Putting the equation of a hyperbola in its standard form is crucial for identifying its characteristics and sketching it properly. The standard equation of a horizontal hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), where \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively.

To convert the given equation \(64x^2 - 36y^2 = 2304\) into standard form, follow these steps:

• Divide each term by 2304 to normalize the equation, setting the right-hand side to 1.
• This results in \(\frac{x^2}{36} - \frac{y^2}{64} = 1\).

In this case, \(a^2 = 36\) and \(b^2 = 64\), which means \(a = 6\) and \(b = 8\). Recognizing the standard form allows us to easily extract other information, such as the vertices and asymptotes, and to confirm the hyperbola’s orientation.

Foci of Hyperbola

The foci are special points located inside each branch of a hyperbola. They are instrumental in defining the hyperbola itself, as their positions determine the curve's shape and features.

To calculate the foci for a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), use this relationship: \(c^2 = a^2 + b^2\). Here, \(c\) represents the distance from the center to each focus.

• Given \(a = 6\) and \(b = 8\), we calculate \(c\) as follows:
• \(c^2 = 6^2 + 8^2 = 36 + 64 = 100\), hence \(c = \sqrt{100} = 10\).

The foci are thus located at \((\pm 10, 0)\). These points are critical in understanding how the hyperbola opens and the overall span of its branches.

Vertices of Hyperbola

Vertices are the points on the hyperbola that lie on the line of symmetry. For the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), they are located horizontally from the center at \((\pm a, 0)\).

In our example, since \(a = 6\), the vertices are at \((6, 0)\) and \((-6, 0)\). These vertices represent the widest part of the hyperbola branches stretching along the x-axis. Knowing the vertices helps define the shape of the hyperbola and its orientation on the coordinate plane.