Problem 25
Question
Find the foci for each equation of an ellipse. Then graph the ellipse. $$ \frac{x^{2}}{225}+\frac{y^{2}}{144}=1 $$
Step-by-Step Solution
Verified Answer
The coordinates of the foci for the ellipse identified by the equation \(\frac{x^{2}}{225}+\frac{y^{2}}{144}=1\) are (-9, 0) and (9, 0). The graph of the ellipse is centered at origin (0, 0) with a width of 30 units and height of 24 units, with foci located at (-9, 0) and (9, 0).
1Step 1: Identify a and b
The equation of an ellipse can be written in the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\). Thus, we identify the values for a and b from this equation: \(a^{2}=225\) and \(b^{2}=144\). We take the square root of these to get \(a=15\) and \(b=12\).
2Step 2: Calculate c
We know that the relationship between a, b, and c in an ellipse is given by \(a^{2}=b^{2}+c^{2}\). Solving this for c, we get \(c=\sqrt{a^{2}-b^{2}}\). So, we substitute a and b: \(c=\sqrt{15^{2}-12^{2}} = \sqrt{225-144} = \sqrt{81} = 9\). Therefore, the foci are located 9 units to the left and right of the center of the ellipse.
3Step 3: Determine foci positions
The foci of the ellipse lie on the x-axis, since a > b in this case. Therefore, the coordinates of the foci are (-c, 0) and (c, 0). Substituting c = 9, we get the foci as (-9, 0) and (9, 0).
4Step 4: Graph the ellipse
To graph the ellipse, note that it's centered at origin (0, 0), the semi-major axis is 15 units long along the x-axis, and the semi-minor axis is 12 units long along the y-axis. The width of the ellipse is thus 30 units (from -15 to 15 along x-axis), and the height is 24 units (from -12 to 12 along the y-axis). Mark the foci at (-9, 0) and (9, 0). Draw the ellipse so it touches the x-axis at -15 and 15 and the y-axis at -12 and 12.
Key Concepts
Foci of an EllipseEllipse EquationGraphing Ellipses
Foci of an Ellipse
Within any ellipse, there exist two crucial points called foci (singular: focus). These points hold great importance as they are used to define the shape of the ellipse itself.
Every point on the ellipse has a unique property. The sum of the distances from any point on the ellipse to each of the foci is constant. This makes the foci integral in understanding the geometry of the ellipse.
Every point on the ellipse has a unique property. The sum of the distances from any point on the ellipse to each of the foci is constant. This makes the foci integral in understanding the geometry of the ellipse.
- For an ellipse with a major axis along the x-axis, as in our example, if 225 is greater than 144, the foci are horizontally aligned.
- The formula to find the distance to the foci from the center is given by: \[ c = \sqrt{a^2 - b^2} \]
- In this equation: \( a = 15 \) and \( b = 12 \).
Ellipse Equation
The ellipse is a shape that holds consistent mathematical characteristics, and these are encapsulated by its equation. The standard form of an ellipse equation is: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where:\
In our example, \( a^2 = 225 \) and \( b^2 = 144 \). From these, we retrieve the values \( a = 15 \) and \( b = 12 \). This geometric representation aids in determining not only the span of the ellipse but also aids in identifying the foci.
where:\
- \( x \) and \( y \) are the coordinates of any point on the ellipse.
- \( a \) and \( b \) are the semi-major and semi-minor axes, respectively.
In our example, \( a^2 = 225 \) and \( b^2 = 144 \). From these, we retrieve the values \( a = 15 \) and \( b = 12 \). This geometric representation aids in determining not only the span of the ellipse but also aids in identifying the foci.
Graphing Ellipses
Graphing an ellipse requires plotting the key features identified in the equation. This is quite straightforward once we have the dimensions and the foci of the ellipse.
To start:
Place the foci on the axis of major length. For our ellipse, ensure that the points (-9, 0) and (9, 0) are marked which represents the foci.
This approach highlights the symmetry and balance of the ellipse, facilitating an accurate graph.
To start:
- Identify the center of the ellipse, which in our example is at (0, 0).
- The semi-major axis, measured along the x-axis, is 15 units. Therefore, the ellipse spans from -15 to 15 on this axis.
- The semi-minor axis, checked along the y-axis, is 12 units, extending from -12 to 12.
Place the foci on the axis of major length. For our ellipse, ensure that the points (-9, 0) and (9, 0) are marked which represents the foci.
This approach highlights the symmetry and balance of the ellipse, facilitating an accurate graph.
Other exercises in this chapter
Problem 24
Identify the vertex, the focus, and the directrix of each graph. Then sketch the graph. $$ x=\frac{1}{24} y^{2} $$
View solution Problem 24
For each equation, find the center and radius of the circle. $$ (x+6)^{2}+y^{2}=121 $$
View solution Problem 25
Use the equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) to identify the shape of the graph that results in each case. a. \(A=C=D=0, B \neq 0, F \neq 0\) b. \(A=B
View solution Problem 25
Write the equation of a hyperbola with the given foci and vertices. foci \((0, \pm 2),\) vertices \((0, \pm 1)\)
View solution