/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Identify the conic as a circle o... [FREE SOLUTION] | 91Ó°ÊÓ

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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. \(9 x^{2}+4 y^{2}+36 x-24 y+36=0\)

Short Answer

Expert verified
The given conic is an ellipse with the center at (-2, 1.5), vertices at (-3, 1.5) and (-1, 1.5). No real foci exist because the value calculated for the foci is an imaginary number.

Step by step solution

01

Rearrange Equation into Standard Form

First, write the equation \(9 x^{2}+4 y^{2}+36 x-24 y+36=0\) in standard form. The standard form of the equation of an ellipse with center at \((h, k)\) and major axis along the x-axis is \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\). To achieve this we will complete the square:
02

Step 2:Calculate centers, radius, vertices and foci

The completed square yields \((3x + 6)^2 + (2y -3)^2 = 9\), which simplifies the equation to \((x + 2)^2/1 + (y - 3/2)^2/2.25 = 1\). From this equation, we ascertain that the center of the ellipse is at (-2, 3/2), the semi-major axis is \(\sqrt{1} = 1\), and the semi-minor axis is \(\sqrt{2.25} = 1.5\). The vertices are at \((h \pm a, k)\) or \((-3, 3/2)\), \((-1, 3/2)\), and the foci can be calculated from the formula \(c = \sqrt{a^{2} - b^{2}}\), which yields \(c = \sqrt{1 - (1.5)^2}\) => \(c = \sqrt{-1.25}\), indicating that it's an imaginary number. Hence, foci do not exist in real numbers. The eccentricity of the ellipse is not applicable as ellipse does not have defined eccentricity.
03

Step3:Graph of the Ellipse

Sketch the graph with the center at (-2, 3/2). Draw the major axis with length of 1 unit along the x-axis, and the minor axis with length of 1.5 units along the y-axis. Mark the vertices at (-3, 3/2) and (-1, 3/2). There are no real foci to mark on the graph. The graph represents an ellipse that is elongated vertically.

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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
The standard form of an ellipse is crucial for identifying its key features and graphing it correctly. It is expressed as \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\) for horizontal ellipses, or \(\frac{(y - k)^2}{a^2} + \frac{(x - h)^2}{b^2} = 1\) for vertical ellipses. Here, \(h, k\) denotes the center of the ellipse, \(a\) is the length of the semi-major axis, and \(b\) represents the length of the semi-minor axis.

The equation shows the precise dimensions and location of the ellipse in Cartesian coordinates. Understanding the standard form allows us to extract information such as the length of axes, center, vertices, and the general shape of the ellipse.
Completing the Square
Completing the square is a mathematical technique used to convert a quadratic equation into its standard form. It involves organizing the x and y terms, adding and subtracting certain values to form perfect squares, and thereby simplifying the equation.

  • Group the \(x\) terms and \(y\) terms together.
  • Factor out the coefficients of \(x^2\) and \(y^2\) if they are not 1.
  • Add and subtract the square of half the coefficient of the x and y terms to create perfect square trinomials.
  • Move the constants to the other side of the equation.
Once in standard form, identifying and graphing conic sections like ellipses becomes more manageable.
Ellipse Characteristics
An ellipse has several defining characteristics that can be deduced from its standard form equation. The center gives us the point in the Cartesian plane where the ellipse is symmetrical. The semi-major (\(a\)) and semi-minor (\(b\)) axes determine the ellipse's width and height—the greater these values, the larger the ellipse.

Vertices are the points on the ellipse furthest from the center along the major axis. Foci are points inside the ellipse that are used to define its shape mathematically, located along the major axis, and eccentricity describes the degree of deviation from being circular. In the given exercise, certain attributes such as real foci and eccentricity are not applicable as they are based on the relationship \(a^2 - b^2\) being positive.
Graphing Ellipses
Graphing ellipses begins with identifying the center, axes, and vertices from the standard form equation. The center is marked first; then, the lengths of the semi-major and semi-minor axes are measured along their respective directions. Vertices are plotted at the ends of these axes.

For the vertical ellipse, the semi-major axis is drawn along the y-axis, and for the horizontal ellipse, it's along the x-axis. The final step is to draw a smooth, continuous curve that connects the vertices and hugs the axes, maintaining the shape of an ellipse. Although the foci were not real in the provided exercise, if they were, they would be plotted and used as focal points to draw the ellipse around.
Conic Sections
Conic sections are the curves obtained by intersecting a cone with a plane. These shapes include circles, ellipses, parabolas, and hyperbolas. The position and angle of the intersecting plane determine the type of conic section:

  • A perpendicular cut to the axis produces a circle.
  • An angled cut that intersects both halves of the cone forms an ellipse.
  • A parallel cut to one side of the cone creates a parabola.
  • An angled cut that crosses the cone twice makes a hyperbola.
The standard form equations of conic sections reflect these geometric properties, and recognizing the patterns in these equations aids in classifying and graphing conic sections.

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Most popular questions from this chapter

The first artificial satellite to orbit Earth was Sputnik I (launched by the former Soviet Union in 1957 ). Its highest point above Earth's surface was 947 kilometers, and its lowest point was 228 kilometers (see figure). The center of Earth was at one focus of the elliptical orbit, and the radius of Earth is 6378 kilometers. Find the eccentricity of the orbit.

Find an equation of the tangent line to the parabola at the given point, and find the \(x\) -intercept of the line. \(x^{2}=2 y,(4,8)\)

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The area of the shaded region in the figure is \(A=\frac{8}{3} p^{1 / 2} b^{3 / 2}\). (a) Find the area when \(p=2\) and \(b=4\). (b) Give a geometric explanation of why the area approaches 0 as \(p\) approaches 0 .

Each cable of the Golden Gate Bridge is suspended (in the shape of a parabola) between two towers that are 1280 meters apart. The top of each tower is 152 meters above the roadway. The cables touch the roadway midway between the towers. (a) Draw a sketch of the bridge. Locate the origin of a rectangular coordinate system at the center of the roadway. Label the coordinates of the known points. (b) Write an equation that models the cables. (c) Complete the table by finding the height \(y\) of the suspension cables over the roadway at a distance of \(x\) meters from the center of the bridge. $$\begin{array}{|c|c|}\hline \text { Distance, } \boldsymbol{x} & \text { Height, } \boldsymbol{y} \\\\\hline 0 & \\\100 & \\\250 & \\\400 & \\\500 & \\\\\hline\end{array}$$

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