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Chapter 11: Problem 18

Find the center, vertices, and foci of each ellipse and graph it. $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$

Short Answer

Expert verified
Center: (0, 0); Vertices: (3, 0) and (-3, 0); Foci: (鈭�5, 0) and (-鈭�5, 0)

Step by step solution

01

Identify the Standard Form

The given equation is \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\). Confirm that this matches the standard form of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\).
02

Determine the Values of a and b

From the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\), identify \(a^{2} = 9\) and \(b^{2} = 4\). Calculate \(a\) and \(b\): \(a = 3\), \(b = 2\).
03

Find the Center of the Ellipse

The center of the ellipse in standard form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) is at the origin, \((0, 0)\).
04

Locate the Vertices

The vertices are along the major axis, which is the x-axis in this case, at \((\text{卤}a, 0)\). So the vertices are located at \((3, 0)\) and \((-3, 0)\).
05

Locate the Co-vertices

The co-vertices are along the minor axis, which is the y-axis in this case, at \((0, \text{卤}b)\). So the co-vertices are located at \((0, 2)\) and \((0, -2)\).
06

Calculate the Distance to the Foci

The distance from the center to each focus is given by \(c = \sqrt{a^2 - b^2}\). Calculate \(c\): \(c = \sqrt{9 - 4} = \sqrt{5}\).
07

Locate the Foci

The foci are along the major axis (the x-axis) at \((\text{卤}c, 0)\). So the foci are at \((\text{卤}\sqrt{5}, 0)\).
08

Graph the Ellipse

Plot the center at \( (0, 0) \). Plot the vertices at \( (3, 0) \) and \( (-3, 0) \). Plot the co-vertices at \( (0, 2) \) and \( (0, -2) \). Plot the foci at \( (\text{卤}\sqrt{5}, 0) \). Draw the ellipse passing through these points.

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Key Concepts

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

Standard Form of an Ellipse
To understand ellipses in algebra, we first need to recognize their standard form. The standard form of an ellipse's equation is: \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\). Here: \(x\) and \(y\) are variables
\(a\) and \(b\) are constants that determine the ellipse's size.
A critical check is if the right-hand side of the equation is 1. Without this, the form isn't standard.
If the equation doesn't match this format, rearrange and simplify it, ensuring the denominator (if any) clears to 1.
Vertices of an Ellipse
Vertices are crucial points on an ellipse, located on its longest axis (major axis).
For an ellipse centered at the origin with equation \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), the vertices are \( (\pm a, 0)\).
They demarcate the farthest extents of the ellipse horizontally if the major axis is along the x-axis, or vertically if the major axis is along the y-axis.
In our problem, the equation \(\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1\) reveals \(a = 3\) and \(b = 2\), therefore, the vertices are at \( (3, 0)\) and \( (-3, 0)\).
Foci of an Ellipse
Foci (plural of focus) are essential points within an ellipse influencing its shape. The distance formula to calculate the focal points is \(c = \sqrt{a^{2} - b^{2}}\).
This distance
\(c\) signifies the space from the center to each focus.
Once \(c\) is found, the foci are placed along the major axis at \( (\pm c, 0)\). For our given equation, \(c = \sqrt{3^{2} - 2^{2}} = \sqrt{9 - 4} = \sqrt{5}\).
Thus, the foci are at \(\pm \sqrt{5} \, ( \approx 2.24)\) from the center.
Graphing Ellipses
Graphing an ellipse involves several ordered steps:
  • Plot the center.
  • Mark the vertices and co-vertices.
  • Pinpoint the foci.
  • Draw a smooth curve connecting these points.
For the exercise, the center is at (0,0).
The vertices at (3, 0) and (-3, 0). Co-vertices at (0, 2) and (0, -2).
The foci at \(\pm \sqrt{5}\). Connect these with a smooth curve, ensuring it captures the elliptical shape.
Equation of an Ellipse
Understanding and identifying the equation of an ellipse is foundational. In our example \(\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1\): The equation fits the standard form
\(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\), where \(a^{2} = 9\) and \(b^{2} = 4\). Thus, \(a = 3\) and \(b = 2\).
Knowing the constants helps plot points and understand the ellipse鈥檚 dimensions.
Always verify the equation fits the standard form to simplify finding properties and graphing.

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Most popular questions from this chapter

Sonic Boom Aircraft such as fighter jets routinely go supersonic (faster than the speed of sound). An aircraft moving faster than the speed of sound produces a cone-shaped shock wave that "booms" as it trails the vehicle. The wave intersects the ground in the shape of one half of a hyperbola and the area over which the "boom" is audible is called the "boom carpet." If an aircraft creates a shock wave that intersects the ground in the shape of the hyperbola \(\frac{x^{2}}{484}-\frac{y^{2}}{100}=1\) (units in miles), how wide is the "boom carpet" 32 miles behind the aircraft?

Find the center, transverse axis, vertices, foci, and asymptotes. Graph each equation. \(2 x^{2}-y^{2}=4\)

Nuclear Power Plant Some nuclear power plants utilize "natural draft" cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. Suppose that such a cooling tower has a base diameter of 400 feet, and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. If the diameter at the top of the tower is 300 feet, how tall is the tower?

Show that the graph of an equation of the form $$A x^{2}+C y^{2}+D x+E y+F=0 \quad A \neq 0, C \neq 0$$ where \(A\) and \(C\) are of the same sign, (a) is an ellipse if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F\) is the same sign as \(A\). (b) is a point if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F=0\). (c) contains no points if \(\frac{D^{2}}{4 A}+\frac{E^{2}}{4 C}-F\) is of opposite sign to \(A\)

Find the center, transverse axis, vertices, foci, and asymptotes, Graph each equation. \(\frac{(x-2)^{2}}{4}-\frac{(y+3)^{2}}{9}=1\)
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