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Chapter 10: Problem 55

Determine the eccentricity of the hyperbola. $$x^{2}/9-y^{2} / 16=1$$

Short Answer

Expert verified
The eccentricity of the hyperbola \(x^{2}/9-y^{2} / 16=1\) is \(\frac{5}{3}\).

Step by step solution

01

Identify the center, and the values of a and b

From the given equation, we can identify the center to be (0,0) as there is no h or k value. The value of a squared (a^2) is 9, and the value of b squared (b^2) is 16. So, the values of a and b are as follows: a = 3 (since \(a^2 = 9\)) b = 4 (since \(b^2 = 16\))
02

Substitute the values of a and b into the formula for eccentricity

Now that we have the values of a and b, substitute them into the eccentricity formula: \(e = \sqrt{1 + \frac{b^2}{a^2}}\) \(e = \sqrt{1 + \frac{4^2}{3^2}}\) \(e = \sqrt{1 + \frac{16}{9}}\) Solve the fraction inside the square root: \(e = \sqrt{1 + \frac{16}{9}}\) \(e = \sqrt{\frac{25}{9}}\) Now, find the square root: \(e = \frac{5}{3}\) So, the eccentricity of the given hyperbola is \(\frac{5}{3}\).

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Key Concepts

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

Center of a Hyperbola
For any hyperbola, the center is a key feature. It acts as a reference point for the rest of the graph. In the standard hyperbola equations, the center is denoted by \(h, k\). If these values are not explicitly stated (as seen as in our original exercise), it means the center is at the origin, \(0, 0\). Without an \(h\) or \(k\) in the equation, you can simplify the process of graphing and solving other properties.
Knowing the center allows you to easily find the axes and vertices, which are critical for understanding the shape and orientation.
Hyperbola Equation
A hyperbola equation has a specific standard form. It can come in two variations:
  • Horizontal: \( rac{(x-h)^2}{a^2} - rac{(y-k)^2}{b^2} = 1\)
  • Vertical: \( rac{(y-k)^2}{b^2} - rac{(x-h)^2}{a^2} = 1\)
In both forms, the constants \(a^2\) and \(b^2\) determine the shape and position of the hyperbola. Our given equation \(x^2/9 - y^2/16 = 1\) follows the horizontal form, centered at \(0, 0\). The beauty of hyperbola equations lies in how they balance symmetry and dimension, contributing to their unique open shape.
Eccentricity Formula
Eccentricity measures how "stretched" a hyperbola is. The formal formula is \(e = \sqrt{1 + \frac{b^2}{a^2}}\), where \('e'\) is always greater than 1. For our specific example, the eccentricity is \(\frac{5}{3}\). This value tells you how much the hyperbola differs from a perfect circle. High eccentricity means it's more elongated, just like you see in planetary orbits.
The formula shows the relation between a and b, guiding you in comparing the x and y stretches.
a and b Values in Hyperbolas
In hyperbolas, \(a\) and \(b\) play a pivotal role in defining the scopes. \(a^2\) and \(b^2\) are derived from the equation's denominators. In the equation \(x^2/9 - y^2/16 = 1\), we identify:
  • \(a = 3\), since \(a^2 = 9\)
  • \(b = 4\), since \(b^2 = 16\)
These values not only aid in calculating eccentricity but also in sketching the hyperbola's asymptotes and determining its direction. Understanding \(a\) and \(b\) is fundamental for exploring more complex hyperbolic functions.

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Most popular questions from this chapter

(a) The electrostatic charge distribution consisting of a charge \(q(q-0)\) at the point \([r, 0]\) and a charge \(-q\) at \([r, \pi]\) is called a dipole. The lines of force for the dipole are given by the equations $$r \quad k \sin ^{2} \theta$$ Use a graphing utility to draw the lines of force for \(k=1,2,3\) (b) The equipotential lines (the set of points with equal electric potential) for the dipole are given by the equations $$r^{2}=m \cos \theta$$ Use a graphing utility to draw the equipotential lines for \(m=1,2,3\) (c) Draw the curves \(r=2 \sin ^{2} \theta\) and \(r^{2}=2 \cos \theta\) using the same polar axis. Estimate the \(x y\) coordinates of the points where the two curves intersect.

Find the points \((x, y)\) at which the curve has: (a) a horizontal tangent: (b) a vertical tangent. Then sketch the curve. $$x(t)=\sin 2 t, \quad y(t)=\sin t$$

Find a parametrization $$x \quad x(t), \quad y \quad y(t)\quad t \in[0,1]$$ for the given curve. The parabolic arc \(x=1-y^{2}\) from (0,-1) to (0,1)

(a) Use a graphing utility to draw the curve $$x(t)=t^{2}, \quad y(t)=t^{3}-t \quad t \text { real. }$$ (b) Your drawing should show that the curve has a loop. Use a CAS to estimate the length of the loop. Round off your answer to four decimal places.

Take \(a>0 .\) The curve $$\begin{array}{l}x(\theta)=3 a \cos \theta+a \cos 3 \theta \\\y(\theta)=3 a \sin \theta-a \sin 3 \theta\end{array}$$ is called a hypocycloid. (a) Use a graphing utility to draw the curves with \(a=1\) 2. \(\frac{1}{2}\) (b) Take \(a=1 .\) Find the area enclosed by the curve. (c) Take \(a=1 .\) Set up a definite integral that gives the area of the surface generated by revolving the curve about the \(x\) -axis.
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