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Chapter 10: Problem 46

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

Short Answer

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

Step by step solution

01

Identify the semi-major and semi-minor axes

From the given equation, \(x^2 / 16+y^2 / 25=1\), we can identify that \(a^2=16\) and \(b^2=25\). So, the semi-major axis is \(b=\sqrt{25}=5\) and the semi-minor axis is \(a=\sqrt{16}=4\).
02

Calculate the eccentricity of the ellipse

We have \(b > a\) in this problem, so we will use the formula for eccentricity: \(e = \sqrt{1 - \frac{a^2}{b^2}}\). Now, plug in \(a = 4\) and \(b = 5\): $$ e = \sqrt{1 - \frac{4^2}{5^2}} = \sqrt{1 - \frac{16}{25}} $$
03

Complete the calculation to find the eccentricity

Continue with the calculation from the previous step: $$ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $$ So, the eccentricity \(e\) 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 Semi-Major and Semi-Minor Axes
The semi-major and semi-minor axes are crucial to the shape of an ellipse. These are the 'radii' of the ellipse, measuring half the length of the ellipse's longest and shortest diameters respectively. The semi-major axis runs from the center to the farthest edge and is denoted as 'a'. Conversely, the semi-minor axis, denoted as 'b', stretches from the center to the closest edge.

Using the given equation, \(x^2 / 16 + y^2 / 25 = 1\), we can deduce that \(a^2 = 16\) and \(b^2 = 25\). Hence, \(a = 4\) and \(b = 5\), signifying that the semi-minor axis is 4 units in length while the semi-major axis is 5 units long. This difference in the axes' lengths gives the ellipse its characteristic oval shape.
Deciphering the Ellipse Equation
The equation of an ellipse in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) provides us with a blueprint of this geometric figure on a coordinate plane. Here, \(a\) and \(b\) symbolize the lengths of the semi-major and semi-minor axes respectively. If \(b > a\), the ellipse stretches more along the y-axis, making it taller. Conversely, if \(a > b\), the ellipse is wider, stretching more along the x-axis.

The given equation \(\frac{x^2}{16} + \frac{y^2}{25} = 1\) tells us the ellipse鈥檚 exact shape and orientation on the graph. By rewriting it as \(x^2 + y^2 \frac{16}{25} = 16\), we can see more clearly that \(b\) is associated with the y-coordinate and is greater than \(a\), indicating a taller ellipse.
Calculating Eccentricity
Eccentricity, denoted as \(e\), measures an ellipse's deviation from being a perfect circle. It is a dimensionless number that ranges from 0 (a circle) to just below 1 (a highly stretched ellipse). To calculate it, use the formula \(e = \sqrt{1 - \frac{a^2}{b^2}}\), where \(a\) is the length of the semi-minor axis and \(b\) is the length of the semi-major axis.

For the ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), we determined that \(a = 4\) and \(b = 5\). Plugging these values into the formula, we get \(e = \sqrt{1 - \frac{16}{25}}\). Simplifying the square root, the eccentricity is thus \(\frac{3}{5}\). This indicates that the ellipse is moderately elongated, being neither a circle (\(e = 0\)) nor too elongated (with a high \(e\) value approaching 1).

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Most popular questions from this chapter

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

Calculate \(d^{2} y / d x^{2}\) at the indicated point without eliminating the parameter \(t.\) $$x(t)=\sin ^{2} t, \quad y(t)=\cos t \quad \text { at } \quad I=\frac{1}{4} \pi$$

Use this method to find the point(s) of self-intersection of each of the following curves. $$x(t)=\sin 2 \pi t, \quad y(t)=2 t-t^{2} \quad 1 \subset[0,4]$$

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.

Determine the eccentricity of the ellipse. $$(x+1)^{2} / 169+(y-1)^{2} / 144=1$$
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