Problem 6
Question
Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$y^{2}+9 x^{2}=9$$
Step-by-Step Solution
Verified Answer
Vertices are at \((0, 3)\) and \((0, -3)\); foci are at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).
1Step 1: Simplify the Equation
Start by dividing every term in the equation \( y^2 + 9x^2 = 9 \) by 9 to simplify it to standard form. This gives us \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \). This is the equation of an ellipse in standard form.
2Step 2: Identify the Major and Minor Axis
In \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \), the larger denominator is 9, indicating that the major axis is vertical. So, this is a vertical ellipse.
3Step 3: Find the Vertices
For a vertical ellipse in the form \( \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \), the vertices are at \((0, \pm a)\). Here, \( a^2 = 9 \) so \( a = 3 \), thus the vertices are \((0, 3)\) and \((0, -3)\).
4Step 4: Find the Foci
For a vertical ellipse, the foci are given by \((0, \pm c)\), where \( c^2 = a^2 - b^2 \). Here, \( b^2 = 1 \), so \( c^2 = 9 - 1 = 8 \), leading to \( c = \sqrt{8} = 2\sqrt{2} \). Thus, the foci are at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).
5Step 5: Sketch the Graph
Draw the ellipse centered at the origin with vertices at \((0, 3)\), \((0, -3)\), and endpoints of the minor axis at \((1, 0)\) and \((-1, 0)\). Show the foci inside the ellipse at \((0, 2\sqrt{2})\) and \((0, -2\sqrt{2})\).
Key Concepts
Vertices of EllipseFoci of EllipseStandard Form of Ellipse
Vertices of Ellipse
The vertices of an ellipse are special points located along the major axis of the ellipse. They are the outermost points that span the longest dimension of the ellipse. For the equation \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \), we have a vertical ellipse because the larger denominator (9) is under the \( y^2 \) term. This tells us that the major axis is aligned vertically.
To find the vertices, we use the formula associated with the vertical ellipse: \( (0, \pm a) \). Here, \( a^2 = 9 \), so \( a = 3 \). Thus, the vertices of this ellipse are at \( (0, 3) \) and \( (0, -3) \), marking the highest and lowest points on the ellipse.
These points help in sketching the ellipse as they define its stature along the major axis. Locating the vertices aids in determining the shape's orientation and emphasized measurement.
To find the vertices, we use the formula associated with the vertical ellipse: \( (0, \pm a) \). Here, \( a^2 = 9 \), so \( a = 3 \). Thus, the vertices of this ellipse are at \( (0, 3) \) and \( (0, -3) \), marking the highest and lowest points on the ellipse.
These points help in sketching the ellipse as they define its stature along the major axis. Locating the vertices aids in determining the shape's orientation and emphasized measurement.
Foci of Ellipse
The foci of an ellipse are two important points located along the major axis, much like the vertices, but their positions are found slightly inside, closer to the center. The unique property of an ellipse is that the sum of distances from any point on the ellipse to these two foci is constant.
For a vertical ellipse like our equation \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \), the foci are given by \( (0, \pm c) \). We find \( c \) using the relationship \( c^2 = a^2 - b^2 \). With \( a^2 = 9 \) and \( b^2 = 1 \), we have \( c^2 = 9 - 1 = 8 \) which gives \( c = \sqrt{8} = 2\sqrt{2} \). Therefore, the foci are at \( (0, 2\sqrt{2}) \) and \( (0, -2\sqrt{2}) \).
Understanding the location of the foci is crucial for recognizing the geometric properties of the ellipse, especially when it comes to drawing or identifying additional elements like the focal length.
For a vertical ellipse like our equation \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \), the foci are given by \( (0, \pm c) \). We find \( c \) using the relationship \( c^2 = a^2 - b^2 \). With \( a^2 = 9 \) and \( b^2 = 1 \), we have \( c^2 = 9 - 1 = 8 \) which gives \( c = \sqrt{8} = 2\sqrt{2} \). Therefore, the foci are at \( (0, 2\sqrt{2}) \) and \( (0, -2\sqrt{2}) \).
Understanding the location of the foci is crucial for recognizing the geometric properties of the ellipse, especially when it comes to drawing or identifying additional elements like the focal length.
Standard Form of Ellipse
The standard form of an ellipse's equation helps in identifying key characteristics of the ellipse, such as its alignment, vertices, and foci. The equation is generally given in the form \( \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \) for a vertical ellipse, and \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) for a horizontal ellipse.
In our exercise, dividing the original equation \( y^2 + 9x^2 = 9 \) by 9 gives \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \). This transformation into its standard form immediately reveals that the ellipse is vertical.
The values of \( a^2 = 9 \) and \( b^2 = 1 \) help us find the vertices, the foci, and the nature of the ellipse. It shows us that \( a \) corresponds to the semi-major axis length and \( b \) to the semi-minor axis length. Recognizing these elements through its standard form simplifies the process of drawing and analyzing the ellipse.
In our exercise, dividing the original equation \( y^2 + 9x^2 = 9 \) by 9 gives \( \frac{y^2}{9} + \frac{x^2}{1} = 1 \). This transformation into its standard form immediately reveals that the ellipse is vertical.
The values of \( a^2 = 9 \) and \( b^2 = 1 \) help us find the vertices, the foci, and the nature of the ellipse. It shows us that \( a \) corresponds to the semi-major axis length and \( b \) to the semi-minor axis length. Recognizing these elements through its standard form simplifies the process of drawing and analyzing the ellipse.
Other exercises in this chapter
Problem 6
Change the polar coordinates to rectangular coordinates. (a) \((4,-\pi / 4)\) (b) \((-2,7 \pi / 6)\)
View solution Problem 6
Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$y^{2}-\frac{x^{2}}{1
View solution Problem 6
Exer. \(1-12\) : Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix. $$(x-3)^{2}=\frac{1}{2}(y+1)$$
View solution Problem 7
Find the eccentricity, and classify the conic. Sketch the graph, and label the vertices. $$r=\frac{4}{\cos \theta-2}$$
View solution