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Chapter 6: Problem 57

Find the eccentricity of the ellipse. \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\)

Short Answer

Expert verified
The eccentricity of the ellipse is \(\frac{\sqrt{5}}{3}\)

Step by step solution

01

Identify a and b

From the equation of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we can see that \(a^{2}=9\) and \(b^{2}=4\), which means that \(a=3\) and \(b=2\). Here, \(a > b\), so no need to swap.
02

Calculate Eccentricity

We can substitute the values of a and b into the formula for the eccentricity of an ellipse. This gives us \(e=\sqrt{1-\frac{2^{2}}{3^{2}}}= \sqrt{1-\frac{4}{9}}=\sqrt{\frac{5}{9}}=\frac{\sqrt{5}}{3}\)

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Key Concepts

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

Ellipse Equation
An ellipse is a set of points on a plane such that the sum of the distances from two fixed points, called foci, to each point on the ellipse is constant. This can be visualized as a stretched circle. The standard equation of an ellipse when centered at the origin is:
  • \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Here, \(a\) represents the semi-major axis, and \(b\) is the semi-minor axis of the ellipse. In our given problem, the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) is in this form.In this equation:
  • The semi-major axis is along the y-axis because 9 is greater than 4.
  • Thus, \(a^2 = 9\) (hence, \(a = 3\)) and \(b^2 = 4\) (thus, \(b = 2\)).
Recognizing the coefficients of the squared terms will guide you to determine whether the ellipse elongates horizontally or vertically.
Eccentricity Calculation
Eccentricity is a measure of how much an ellipse deviates from being a circle. For an ellipse, the eccentricity ranges from 0 to 1. A lower eccentricity means the ellipse is closer to being a circle; a higher value indicates it's more stretched.The formula to calculate the eccentricity \(e\) of an ellipse is:
  • \[ e = \sqrt{1 - \frac{b^2}{a^2}} \]
Given, \(a = 3\) and \(b = 2\) from the equation
  • \[ e = \sqrt{1 - \frac{2^2}{3^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]
This simple calculation tells us that the ellipse is moderately stretched since \(e\) is greater than 0 but less than 1.
Conic Sections
Conic sections are the curves obtained by intersecting a cone with a plane. They are fundamental in geometry and come in four types: circle, ellipse, parabola, and hyperbola.An ellipse is one of the conic sections, formed when the plane intersects the cone at an angle that is inclined to its base but doesn鈥檛 intersect the base itself.In general:
  • A circle is a special type of ellipse with an eccentricity of 0.
  • An ellipse has an eccentricity between 0 and 1, as seen in our calculation \(\frac{\sqrt{5}}{3}\).
  • A parabola has an eccentricity of 1, and a hyperbola has an eccentricity greater than 1.
Understanding these categories can help you predict the shape and properties of the curve based solely on its equation.

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Most popular questions from this chapter

Find the vertex, focus, and directrix of the parabola. Use a graphing utility to graph the parabola. \(x^{2}+4 x+6 y-2=0\)

Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(x^{2}+5 y^{2}-8 x-30 y-39=0\)

Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Vertical axis and passes through the point (-3,-3)

Sketch the graph of the ellipse, using latera recta. \(\frac{x^{2}}{4}+\frac{y^{2}}{1}=1\)

Identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(\frac{x^{2}}{64}+\frac{y^{2}}{28}=1\)
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