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Chapter 11: Problem 24

In Exercises \(17-24\) , find the eccentricity of the hyperbola. Then find and graph the hyperbola's foci and directrices. $$64 x^{2}-36 y^{2}=2304$$

Short Answer

Expert verified
Eccentricity is \(\frac{5}{3}\), foci are \((10, 0)\) and \((-10, 0)\), and directrices are \(x = 3.6\) and \(x = -3.6\).

Step by step solution

01

Identify the Standard Form

The given equation is \(64x^2 - 36y^2 = 2304\). A hyperbola in standard form is given by \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). To start, divide every term of the equation by 2304 to transform the hyperbola into its standard form.
02

Simplify the Equation

Divide the entire equation \(64x^2 - 36y^2 = 2304\) by 2304 to get \(\frac{64x^2}{2304} - \frac{36y^2}{2304} = 1\). Simplifying gives \(\frac{x^2}{36} - \frac{y^2}{64} = 1\). Thus, it is in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) with \(a^2 = 36\) and \(b^2 = 64\).
03

Calculate the Eccentricity

For a hyperbola, the eccentricity \(e\) is given by \(e = \sqrt{1 + \frac{b^2}{a^2}}\). Substitute \(a^2 = 36\) and \(b^2 = 64\) into the formula to get \(e = \sqrt{1 + \frac{64}{36}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}\).
04

Find the Foci

The foci of the hyperbola are found using the formula \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). So, \(c = \sqrt{36 + 64} = \sqrt{100} = 10\). Thus, the foci are \((\pm 10, 0)\), which means \((10, 0)\) and \((-10, 0)\).
05

Determine the Directrices

The directrices of the hyperbola are given by \(x = \pm \frac{a^2}{c}\). From Step 4, we have \(a^2 = 36\) and \(c = 10\). Therefore, \(x = \pm \frac{36}{10} = \pm 3.6\). Thus, the equations of the directrices are \(x = 3.6\) and \(x = -3.6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eccentricity of a Hyperbola
The eccentricity of a hyperbola is a crucial parameter that determines its shape. Simply put, it tells you how "stretched" the hyperbola is. For hyperbolas, the eccentricity value is always greater than one.
To find the eccentricity of a hyperbola, we use the formula:
  • \( e = \sqrt{1 + \frac{b^2}{a^2}} \)
In our example, we identified from the standard form equation \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \) that \( a^2 = 36 \) and \( b^2 = 64 \).
Plugging these values into the formula gives us:
  • \( e = \sqrt{1 + \frac{64}{36}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \)
This result indicates the hyperbola is more elongated, as an eccentricity of \(\frac{5}{3}\) shows a substantial deviation from being a circle, which would have an eccentricity of zero.
Foci of a Hyperbola
The foci are two distinct points that are extremely important in the definition and geometry of a hyperbola. They lie along the transverse axis of the hyperbola.
These points serve as a guide to understand how the hyperbola curves. The distance between each point on the hyperbola and the foci maintains a constant relationship.
To find the foci, use:
  • \( c = \sqrt{a^2 + b^2} \)
For our equation, we have \( a^2 = 36 \) and \( b^2 = 64 \). Calculating \( c \) gives
  • \( c = \sqrt{36 + 64} = \sqrt{100} = 10 \)
Thus, the foci are located at \( (\pm 10, 0) \), which means the two foci are directly at \(10\) units to the left and right of the origin along the x-axis.
Directrices of a Hyperbola
Directrices are lines that are used to define the shape and position of a hyperbola. They are equally important as the foci when considering the foundational properties of a hyperbola.
Each hyperbola has two directrices, and they are located perpendicular to the transverse axis at specific distances from the origin. The mathematical expression for the directrices related to the hyperbola in our context is:
  • \( x = \pm \frac{a^2}{c} \)
We previously calculated \( a^2 = 36 \) and \( c = 10 \). Substituting these into the formula gives:
  • \( x = \pm \frac{36}{10} = \pm 3.6 \)
Hence, the directrices are the vertical lines \( x = 3.6 \) and \( x = -3.6 \), running parallel to the y-axis at a distance of \( 3.6 \) units from the origin.

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Most popular questions from this chapter

Volume The region bounded on the left by the \(y\) -axis, on the right by the hyperbola \(x^{2}-y^{2}=1,\) and above and below by the lines \(y=\pm 3\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid.

If lines are drawn parallel to the coordinate axes through a point \(P\) on the parabola \(y^{2}=k x, k>0,\) the parabola partitions the rectangular region bounded by these lines and the coordinate axes into two smaller regions, \(A\) and \(B .\) $$ \begin{array}{l}{\text { a. If the two smaller regions are revolved about the } y \text { -axis, show }} \\ {\text { that they generate solids whose volumes have the ratio } 4 : 1 .} \\ {\text { b. What is the ratio of the volumes generated by revolving the }} \\ {\text { regions about the } x \text { -axis? }}\end{array} $$

Most centroid calculations for curves are done with a calculator or computer that has an integral evaluation program. As a case in point, find, to the nearest hundredth, the coordinates of the centroid of the curve $$ x=t^{3}, \quad y=3 t^{2} / 2, \quad 0 \leq t \leq \sqrt{3} $$

Exercises \(35-38\) give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation from the information given. $$ \begin{array}{l}{\text { Foci: }(0, \pm \sqrt{2})} \\ {\text { Asymptotes: } y=\pm x}\end{array} $$

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=8 y, \quad \text { right } 1, \text { down } 7 $$
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