Problem 30 - Thomas Calculus

Question

In Exercises \(23-30\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$ 64 x^{2}-36 y^{2}=2304 $$

Step-by-Step Solution

Answer

Eccentricity is \(\frac{5}{3}\). Foci: \((\pm 10, 0)\). Directrices: \(x = \pm 3.6\).

Step 1: Rewrite in Standard Form

The standard form of a hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Start by dividing every term in the given equation by 2304: \( \frac{64x^2}{2304} - \frac{36y^2}{2304} = 1 \). Simplifying gives us \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \). Now the equation is in standard form where \( a^2 = 36 \) and \( b^2 = 64 \).

Step 2: Calculate Eccentricity

For a hyperbola, the eccentricity \( e \) is defined as \( e = \sqrt{1 + \frac{b^2}{a^2}} \). From the standard form, \( a^2 = 36 \) and \( b^2 = 64 \). Thus, \( \frac{b^2}{a^2} = \frac{64}{36} = \frac{16}{9} \). Therefore, \( e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \).

Step 3: Calculate Foci

The foci of the hyperbola are located at \( (\pm c, 0) \), where \( c = a \times e \). Since \( a = \sqrt{36} = 6 \) and \( e = \frac{5}{3} \), we find \( c = 6 \times \frac{5}{3} = 10 \). Thus, the foci are at \( (\pm 10, 0) \).

Step 4: Calculate Directrices

The directrices are lines given by \( x = \pm \frac{a}{e} \). Using \( a = 6 \) and \( e = \frac{5}{3} \), we have \( \frac{a}{e} = \frac{6}{\frac{5}{3}} = \frac{18}{5} = 3.6 \). Therefore, the equations of the directrices are \( x = \pm 3.6 \).

Step 5: Graph the Hyperbola

To graph the hyperbola \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \), plot its center at the origin. Use \( a = 6 \) and \( b = 8 \) to mark the vertices at \( (\pm 6, 0) \) and the \( y \) extents at \( (0, \pm 8) \). The foci at \( (\pm 10, 0) \) and directrices \( x = \pm 3.6 \) can also be marked.

Key Concepts

Eccentricity

The eccentricity of a hyperbola is a crucial parameter that tells us how much the conic section deviates from being circular.
In simple terms, it measures the "ovalness" of the hyperbola.
Eccentricity, usually denoted as \( e \), is defined for a hyperbola by the formula:

• \( e = \sqrt{1 + \frac{b^2}{a^2}} \)

Here, \( a^2 \) and \( b^2 \) are derived from the hyperbola’s standard form.
Specifically, \( a^2 = 36 \) and \( b^2 = 64 \), we get:

• \( \frac{b^2}{a^2} = \frac{64}{36} = \frac{16}{9} \)
• Thus, \( e = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \)

Eccentricity greater than 1 is a definitive indication of a hyperbola.

Foci

The foci of a hyperbola are two special points that are essential for its definition.
They are located along the transverse axis, outside the vertices, and are symmetric about the center.
To find the foci, we use the formula:

• \( c = a \times e \)

In our example:

• \( a = 6 \) and the previously calculated \( e = \frac{5}{3} \)
• Thus, \( c = 6 \times \frac{5}{3} = 10 \)

Therefore, the foci are located at \( (\pm 10, 0) \).
These points are used when defining the hyperbola's reflective properties.

Directrices

Directrices are linear guides that play an essential role in a hyperbola’s construction.
Every hyperbola is associated with two directrices, which are vertical lines parallel to the transverse axis.
The directrices are given by the equation:

• \( x = \pm \frac{a}{e} \)

For our hyperbola:

• \( a = 6 \) and \( e = \frac{5}{3} \)
• Thus, \( \frac{a}{e} = \frac{6}{\frac{5}{3}} = \frac{18}{5} = 3.6 \)

Thus, the directrices are the lines \( x = \pm 3.6 \).
These act as boundary markers for determining points that satisfy the hyperbola's equation.

Standard Form of Hyperbola

The standard form of a hyperbola is fundamental for identifying its geometrical properties.
It rests on the expression:

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

This gives a clear idea of how the hyperbola is oriented in the plane.
For our exercise, the original equation was \( 64x^2 - 36y^2 = 2304 \).

• Dividing each term by 2304 puts it into standard form: \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \)

In this standard form, \( a^2 = 36 \) and \( b^2 = 64 \) identifying \( a = 6 \) and \( b = 8 \).
This formulation helps in calculating other properties like eccentricity, foci, and directrices.