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Determine the eccentricity of the ellipse. $$x^{2} / 25+y^{2} / 16=1$$

Short Answer

Expert verified
The eccentricity of the ellipse given by the equation \(\frac{x^2}{25} + \frac{y^2}{16} = 1\) is \(\frac{3}{5}\).

Step by step solution

01

Identify the values of a and b

In the given equation, \(\frac{x^2}{25} + \frac{y^2}{16} = 1\), we can see that a^2 = 25 and b^2 = 16. Therefore, taking the square root of each of these values, we have: - a = 5 - b = 4
02

Use the eccentricity formula

Now that we have the values of a and b, we can use the eccentricity formula: \(e = \sqrt{1 - \frac{b^2}{a^2}}\)
03

Plug in the values of a and b

Replacing a and b in the formula, we get: \(e = \sqrt{1 - \frac{4^2}{5^2}}\)
04

Simplify the expression

Calculate the value of the expression inside the square root: \(e = \sqrt{1 - \frac{16}{25}}\)
05

Calculate the eccentricity

Subtract the fraction from 1 and compute the square root: \(e = \sqrt{\frac{9}{25}}\) \(e = \frac{3}{5}\) The eccentricity of the ellipse is \(\frac{3}{5}\).

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Key Concepts

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

Understanding the Ellipse Equation
The equation of an ellipse in its standard form can be expressed as \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This is the basic equation that represents the way an ellipse is stretched across the Cartesian plane. The variables \(a\) and \(b\) are significant as they denote the semi-major and semi-minor axes respectively. These axes can be thought of as the radius of the ellipse in the horizontal and vertical directions.
In the provided exercise, the ellipse equation is expressed as \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \). By comparing this with the standard form, we find that \(a^2 = 25\) and \(b^2 = 16\).
  • The semi-major axis, \(a\), is along the x-direction since \(a^2\) is larger than \(b^2\).
  • The semi-minor axis, \(b\), is along the y-direction.
Therefore, \(a = 5\) and \(b = 4\). This foundational understanding of the ellipse equation helps in further calculations like determining the eccentricity of the ellipse.
Eccentricity Formula Explained
Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. It is crucial in conic sections and indicates the roundness of the ellipse. The eccentricity of an ellipse is calculated using the formula: \( e = \sqrt{1 - \frac{b^2}{a^2}} \). Here, \(e\) is the eccentricity, \(b\) represents the semi-minor axis, and \(a\) represents the semi-major axis.
The formula essentially accounts for the ratio of the axes. If the value of \(e\) is closer to 0, the ellipse is more circular. Conversely, if \(e\) is closer to 1, the ellipse is more elongated.
In our case, substituting \(b = 4\) and \(a = 5\) into the formula given in the problem, we calculate:
  • Complete the square inside: \( \frac{4^2}{5^2} = \frac{16}{25} \).
  • Compute \(e\): \( e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \).
Thus, the eccentricity is \(\frac{3}{5}\). This implies that the ellipse is moderately elongated.
Insights into Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. The main types of conic sections are circles, ellipses, parabolas, and hyperbolas. Among these, the ellipse is particularly fascinating due to its properties and applications.
The ellipse, as a conic section, can be visualized as an oval shape formed when the intersection plane cuts through the cone at an angle to its base that is not perpendicular. The defining characteristic of an ellipse comes from its equation and properties, particularly the eccentricity.
  • A circle is a special case where the eccentricity \(e = 0\).
  • An ellipse has an eccentricity \(0 < e < 1\).
  • Ellipses are used in planetary orbits, optics, and architecture due to their unique properties.
Understanding conic sections, especially ellipses, provides us insight into various natural and engineered systems that utilize their geometric properties.

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Most popular questions from this chapter

Find a parametrization $$x \quad x(t), \quad y \quad y(t)\quad t \in[0,1]$$ for the given curve. The curve \(y^{2}=x^{3}\) from (4,8) to (1,1)

Verify that \(x^{\prime}(0)=y^{\prime}(0)=0\) and that the given description holds at the point where \(t=0 .\) Sketch the curve. $$x(t)=t^{5}, \quad y(t)=t^{3} ; \quad \text { vertical tangent. }$$

Assume that the curve $$C: x=x(t), \quad y \equiv y(t) \quad t \in[c, d]$$ is the graph of a nonnegative function \(y=f(x)\) over an interval \([a, b] .\) Assume that \(x^{\prime}(t)\) and \(y(t)\) are continuous, \(x(c): a\) and \(x(d)=b\) (The area under a parametrized curve) Show that HINT: since \(C\) is the graph of \(f, y(t)=f(x(t))\)

Let \(C\) be a simple curve in the upper half-plane parameterized by a pair of continuously differentiable functions. $$x=x(t), \quad y=y(t) \quad t \in[c, d]$$ By revolving \(C\) about the \(x\) -axis, we obtain a surface of revolution, the area of which we the by symmetry, the centroid of the surface lies on the \(x\) axis. Thus the centroid is completely determined by its \(x\) -coordinate \(\bar{x}\). Show that by assuming the following additivity principle: if the surface is broken up into \(n\) surfaces of revolution with areas \(A_{1} \ldots \ldots A_{n}\) and the centroids of the surfaces have \(x\) -coordinates \(\bar{x}_{1} \ldots \ldots \bar{x}_{n},\) then \(\bar{X} A=\bar{x}_{1} A_{1}+\cdots+\bar{x}_{n} A_{n}\).

(a) Use a graphing utility to draw the curves $$r=1+\sin \theta \quad \text { and } \quad r^{2}=4 \sin 2 \theta$$ using the same polar axis. (b) Use a CAS to find the points where the two curves intersect.

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