Problem 41
Question
In Exercises 29-52, identify the conic as a circle or an ellipse. Then find the center, radius, vertices, foci, and eccentricity of the conic (if applicable), and sketch its graph. \(9x^2+4y^2+36x-24y+36=0\)
Step-by-Step Solution
Verified Answer
The given equation forms a circle with center at (-2,3), radius 1, vertices at (-3,3) and (-1,3), foci at (-2,3) and eccentricity of 0.
1Step 1: Rewrite in standard form
Group the x and y terms and complete the square to rewrite the equation in standard form: Dividing entire equation by 4 would give us \(2.25x^2 + 4x + 9y^2 - 6y +9 = 0\)which can be rewritten as: \((x + 2)^2 + 1y^2 - 6y +9 = 0\) which can be further simplified to:\((x + 2)^2 + (y - 3)^2 = 1 \)
2Step 2: Identify the conic and find the center
The equation \((x + 2)^2 + (y - 3)^2 = 1 \) is in the standard form of a circle, where the center is given by (-2,3). The coefficient of x and y are both 1, indicating that this conic section is indeed a circle.
3Step 3: Radius, Vertices and Foci
For a circle, the radius, \(r\), is given by \(\sqrt{h}\) where \(h\) is the constant on the right side of the equation.Thus, radius, r = \(\sqrt{1} = 1\)Vertices of a circle are endpoints of the diameter. Therefore our vertices can be found at (-2-1,3) and (-2+1,3) or (-3,3) and (-1,3)For a circle, the foci coincide with the center, thus foci is at (-2,3)
4Step 4: Eccentricity of the conic
Since it's a circle, the eccentricity is 0 as the two foci coincide.
5Step 5: Sketch the graph
Plot the center, vertices, and foci on a graph. Draw a circle with a radius of 1. This gives us a sketch of the conic.
Key Concepts
CircleEllipseEccentricityVerticesFoci
Circle
A circle is one of the simplest forms of conic sections. It is defined as the set of all points in a plane that are equidistant from a fixed point, known as the center. The standard equation for a circle is \[ (x - h)^2 + (y - k)^2 = r^2 \] where
Circles are unique among conic sections due to their perfect symmetry and constant radius.
- \((h, k)\) is the center of the circle
- \(r\) is the radius
Circles are unique among conic sections due to their perfect symmetry and constant radius.
Ellipse
An ellipse resembles a stretched circle and can be described as the set of all points for which the sum of the distances from two fixed points (foci) is constant. Though similar in appearance to a circle, an ellipse has different lengths for major and minor axes. The standard form of the ellipse’s equation is \[ \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \] where:
- \((h, k)\) is the center
- \(a\) is the semi-major axis
- \(b\) is the semi-minor axis
Eccentricity
Eccentricity helps us determine how much a conic section deviates from being circular. It is represented by the symbol \(e\). For different conic sections, the eccentricity varies:
- Circle: \(e = 0\)
- Ellipse: \(0 < e < 1\)
- Parabola: \(e = 1\)
- Hyperbola: \(e > 1\)
Vertices
Vertices in conic sections are key points that help in defining the shape and extent of the figure. For a circle, vertices are typically the endpoints of the diameter. In our circle, with a center at \((-2, 3)\) and a radius of 1, the vertices are at the points \((-3, 3)\) and \((-1, 3)\). These points represent the farthest edges along a straight line passing through the center.
- For a circle: Defined by its diameter.
- For an ellipse: Located at either end of the major axis.
Foci
Foci are specific points in conic sections from which distances help define the shape's properties.
- In a circle, the foci coincide with the center, making a circle unique as it involves only one central point, \((-2, 3)\) in this exercise.
- In an ellipse, the foci are two distinct points located along the major axis.
Other exercises in this chapter
Problem 41
In Exercises 41-48, find a set of parametric equations for the rectangular equation using (a) \(t=x\) and (b) \(t=2-x\). \(y=3x-2\)
View solution Problem 41
In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((0, 4), (0, 0); \quad\) passes through t
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In Exercises 33-46, find the vertex, focus, and directrix of the parabola, and sketch its graph. \((x+3)^2 = 4(y-\frac{3}{2})\)
View solution Problem 41
In Exercises 37-46, find the angle \(\theta\) (in radians and degrees)between the lines. \(x - 2y = 7\) \(6x + 2y = 5\)
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