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Chapter 10: Problem 47

Determine the eccentricity of the ellipse. $$(x-1)^{2} / 25+(y+2)^{2} / 9=1$$

Short Answer

Expert verified
The eccentricity of the ellipse \(\frac{(x-1)^2}{25} + \frac{(y+2)^2}{9} = 1\) can be calculated using the formula \(e = \sqrt{1 - \frac{b^2}{a^2}}\). By comparing the given equation with the standard equation of an ellipse, we find the semi-major axis \(a = 5\) and the semi-minor axis \(b = 3\). Plugging these values into the formula, we get the eccentricity \(e = \sqrt{1 - \frac{3^2}{5^2}} = \frac{4}{5}\).

Step by step solution

01

Find the semi-major and semi-minor axes lengths

The given equation of ellipse is \(\frac{(x-1)^2}{25} + \frac{(y+2)^2}{9} = 1\). We can see that this is the standard equation of an ellipse, which is \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Comparing the given equation with the standard equation, we have \(a^2 = 25\) and \(b^2 = 9\). Thus, the semi-major axis is \(a = \sqrt{25} = 5\) and the semi-minor axis is \(b = \sqrt{9} = 3\).
02

Calculate the eccentricity

Now that we have the lengths of the semi-major axis (a) and the semi-minor axis (b), we can calculate the eccentricity of the ellipse using the formula \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Plug in the values of a and b: \(e = \sqrt{1 - \frac{3^2}{5^2}} = \sqrt{1 - \frac{9}{25}}= \sqrt{\frac{16}{25}}\) We have the eccentricity e as \(\frac{4}{5}\). So, the eccentricity of the given ellipse is \(\frac{4}{5}\).

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Key Concepts

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

Ellipse
An ellipse is a shape that looks like a flattened circle. It is defined as the set of all points where the sum of the distances from two fixed points, known as the foci, is constant. This shape is very common and can be seen in everyday objects like racetracks or even the orbits of planets. Ellipses have different properties:
  • They are symmetric along both of their axes.
  • The longest axis is called the "major axis" and the shorter one is the "minor axis."
  • Each axis has a corresponding semi-axis, which is half of the length of the full axis.
Understanding an ellipse involves recognizing these properties and their relationships.
Eccentricity
Eccentricity measures how much an ellipse deviates from being a perfect circle. It is denoted as "e" and is a number between 0 and 1 for ellipses.Here's how it works:
  • If the eccentricity is 0, the ellipse is a circle, meaning both axes are equal.
  • If it's closer to 1, the ellipse is elongated, like an oval.
  • The formula used to calculate eccentricity is:
    \( e = \sqrt{1 - \frac{b^2}{a^2}} \)
    where \( a \) is the semi-major axis, and \( b \) is the semi-minor axis.
By knowing the axes, eccentricity helps determine the shape's exact form.
Semi-major axis
The semi-major axis is one of the most significant measures in an ellipse. It represents half of the longest diameter from one side of the ellipse to the other.Key points about the semi-major axis:
  • Denoted by \( a \), it dictates the widest point of the ellipse.
  • It plays a central role in calculating the eccentricity.
  • The length of the semi-major axis is found using half of the full major axis measurement.
In the standard ellipse equation, it is connected to the variable under the \( x^2 \) term when \( a > b \).
Semi-minor axis
The semi-minor axis complements the semi-major axis by being the shorter half of the diameter in an ellipse.Details about the semi-minor axis:
  • Represented by \( b \), it's shorter than the semi-major axis for a typical ellipse, unless the shape is a circle.
  • It's crucial for calculating properties like eccentricity.
  • It impacts the height of the ellipse depending on its orientation.
Within the ellipse equation, the variable linked with the \( y^2 \) term provides the measurement for the semi-minor axis when \( a > b \).

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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 \(y^{3}= x^{2}\) from (1,1) to (8,4)

Find the a:ea of the surface generated by revolving the curve about the \(x\) -axis. \(f(x)=\frac{1}{3} x^{3}, \quad x \in[0,2]\).

Find the points \((x, y)\) at which the curve has: (a) a horizontal tangent: (b) a vertical tangent. Then sketch the curve. $$x(t)=3 t-t^{3}, \quad y(t)=t+1$$

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

Find the a:ea of the surface generated by revolving the curve about the \(x\) -axis. \(y=\cos x . x \in\left[0, \frac{1}{2} \pi\right]\).
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