Problem 16
Question
Sketch the graph of each equation. \(\frac{(x-3)^{2}}{9}+\frac{(y+3)^{2}}{16}=1\)
Step-by-Step Solution
Verified Answer
Sketch the ellipse with center (3, -3); vertices at (3, 1), (3, -7), (6, -3), and (0, -3).
1Step 1: Identify the Equation Type
The given equation \(\frac{(x-3)^{2}}{9}+\frac{(y+3)^{2}}{16}=1\) is the standard form of an ellipse. The general form of an ellipse can be written as \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where \((h, k)\) is the center of the ellipse.
2Step 2: Determine the Center of the Ellipse
From the equation \(\frac{(x-3)^2}{9}+\frac{(y+3)^2}{16}=1\), compare it with the general form to find \(h = 3\) and \(k = -3\). Thus, the center of the ellipse is \((3, -3)\).
3Step 3: Identify the Semi-major and Semi-minor Axes
Compare the denominators in the ellipse equation: \(9 = a^2\) and \(16 = b^2\). This implies \(a = 3\) and \(b = 4\). Since \(b > a\), the semi-major axis is along the y-axis with length 4, and the semi-minor axis is along the x-axis with length 3.
4Step 4: Find the Vertices from the Center
The vertices along the semi-major axis (y-axis) are located \(b\) units away from the center: \((3, -3 + 4) = (3, 1)\) and \((3, -3 - 4) = (3, -7)\). The vertices along the semi-minor axis (x-axis) are: \((3 + 3, -3) = (6, -3)\) and \((3 - 3, -3) = (0, -3)\).
5Step 5: Sketch the Ellipse
Plot the center of the ellipse at \((3, -3)\). Mark the vertices at \((3, 1), (3, -7), (6, -3), (0, -3)\). Drawing an oval shape connecting the vertices will form the ellipse with the major axis aligned vertically.
Key Concepts
Conic SectionsGraphingSemi-major and Semi-minor AxesVerticesCenter of Ellipse
Conic Sections
Conic sections are curves that are created by intersecting a plane with a cone. These include ellipses, circles, parabolas, and hyperbolas. Today, let's specifically talk about ellipses. An ellipse appears as an elongated circle and is one of the most important shapes in this category.
When a plane intersects a cone at an angle that is oblique but not parallel to the base of the cone, and doesn't cut through both nappes (upward and downward stretches of the cone), an ellipse is formed. This set of points maintains the property where the sum of the distances from any point on the ellipse to two fixed points, called foci, remains constant. This elegant mathematical property underpins the beautiful shape of ellipses we often see in various natural and human-made structures.
When a plane intersects a cone at an angle that is oblique but not parallel to the base of the cone, and doesn't cut through both nappes (upward and downward stretches of the cone), an ellipse is formed. This set of points maintains the property where the sum of the distances from any point on the ellipse to two fixed points, called foci, remains constant. This elegant mathematical property underpins the beautiful shape of ellipses we often see in various natural and human-made structures.
Graphing
Graphing an ellipse involves translating the theoretical equation into a visual representation on a coordinate plane. It all starts with understanding the equation's standard form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] Here,
- \((h, k)\) is the center of the ellipse.
- \(a\) represents the semi-major axis' length.
- \(b\) describes the semi-minor axis' length.
Semi-major and Semi-minor Axes
In any ellipse, the semi-major axis is the longest diameter running through the center from one side of the ellipse to the opposite side. The semi-minor axis is the shortest diameter, perpendicular to the semi-major axis at the center. It's important to identify these axes because they determine the length and orientation of the ellipse.
In our given equation, \[\frac{(x-3)^2}{9} + \frac{(y+3)^2}{16} = 1\] we find that \(a = 3\) and \(b = 4\). Because \(b > a\), the semi-major axis is vertical. Therefore, the semi-major axis stretches along the y-axis, while the semi-minor axis spans the x-axis. Knowing which axis is major helps determine how elongated your ellipse's shape is and indicates the direction in which it stretches.
In our given equation, \[\frac{(x-3)^2}{9} + \frac{(y+3)^2}{16} = 1\] we find that \(a = 3\) and \(b = 4\). Because \(b > a\), the semi-major axis is vertical. Therefore, the semi-major axis stretches along the y-axis, while the semi-minor axis spans the x-axis. Knowing which axis is major helps determine how elongated your ellipse's shape is and indicates the direction in which it stretches.
Vertices
Vertices are key points that help in sketching the full shape of an ellipse. For an ellipse, vertices are located at the ends of both the semi-major and semi-minor axes.
Analyzing our ellipse: Centered at \((3, -3)\) with semi-major and semi-minor axes lengths of 4 and 3 units, respectively, we identify the vertices as follows:
Analyzing our ellipse: Centered at \((3, -3)\) with semi-major and semi-minor axes lengths of 4 and 3 units, respectively, we identify the vertices as follows:
- Along the semi-major axis (y-axis): \((3, 1)\) and \((3, -7)\)
- Along the semi-minor axis (x-axis): \((6, -3)\) and \((0, -3)\)
Center of Ellipse
The center of an ellipse serves as the origin point, around which the ellipse is symmetrically distributed. It's crucial to locate the center accurately as it guides the placement of other significant features like the axes and vertices.
From the equation \(\frac{(x-3)^2}{9} + \frac{(y+3)^2}{16} = 1\), we can directly find that the center is at the point \((3, -3)\). Identifying the center helps set up the framework for graphing the ellipse, ensuring that each part of the ellipse is evenly spaced around this central point.
With this understanding, you can confidently graph the oval shape, knowing it is symmetrical relative to its center.
From the equation \(\frac{(x-3)^2}{9} + \frac{(y+3)^2}{16} = 1\), we can directly find that the center is at the point \((3, -3)\). Identifying the center helps set up the framework for graphing the ellipse, ensuring that each part of the ellipse is evenly spaced around this central point.
With this understanding, you can confidently graph the oval shape, knowing it is symmetrical relative to its center.
Other exercises in this chapter
Problem 15
The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$y=-3(x-1)^{2}+5$$
View solution Problem 16
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {x^{2}+y^{2}=9} \\ {x+y=5} \end{array}\right. $$
View solution Problem 16
The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$y=-4(x-2)^{2}+2$$
View solution Problem 17
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{array}{l} {y=x^{2}-4} \\ {y=x^{2}-4 x} \end{array}\right. $$
View solution