Problem 48
Question
Find the eccentricity of the ellipse. $$4 x^{2}+3 y^{2}-8 x+18 y+19=0$$
Step-by-Step Solution
Verified Answer
The eccentricity of the ellipse is \(\sqrt{2/3}\)
1Step 1: Rewrite the equation in standard form
Divide the given equation by 4 and complete the squares to move it into a form that resembles the standard equation for an ellipse. The equation becomes: \(4(x - 1)^2/4 + 9(y + 1)^2/12 = 1\)
2Step 2: Identify the center and semi-axes
From the standard form, we can find the center of the ellipse is at (1, -1), and \(a^2=4, b^2=12\). So, \(a=2\) is the semi minor axis and \(b=\sqrt{12}\) is the semi major axis.
3Step 3: Compute the eccentricity
Substitute \(a\) and \(b\) into the formula \(\sqrt{1 - (a^2 / b^2)}\) which yields \(e = \sqrt{1 - (4/12)} = \sqrt{2/3}\).
Key Concepts
Standard Form of an EllipseCompleting the SquareEllipse CenterSemi-major and Semi-minor Axes
Standard Form of an Ellipse
To solve problems involving ellipses, understanding their equation's standard form is essential. The standard form of an ellipse's equation provides a clear framework, written as \[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\]where
- \((h, k)\) represents the center of the ellipse.
- \(a\) and \(b\) are the lengths of the semi-axes.
Completing the Square
Completing the square is a vital algebraic technique used to transform a quadratic expression into a perfect square trinomial. This process is crucial for rewriting the equation of an ellipse in standard form.Let's look at how completing the square helps with an equation like \[4x^2 + 3y^2 - 8x + 18y + 19 = 0.\]
- First, separate the equation into parts containing \(x\) and \(y\): \(4(x^2 - 2x) + 3(y^2 + 6y) = -19\).
- To complete the square for \(x\), take half of \(-2\) (which is \(-1\)), square it, and adjust the equation: \((x - 1)^2\). Similarly, for \(y\), take half of \(6\) (which is \(3\)), square it, and adjust the term: \((y + 3)^2\).
Ellipse Center
The center of an ellipse is a critical concept in defining its position and ensuring a proper understanding of its geometry. In the ellipse's standard form equation:\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1,\]\((h, k)\) represents the ellipse center. Identifying this point provides insights into the symmetry and orientation of the ellipse relative to the Cartesian coordinate system.For example, if you successfully convert an equation like \[4(x - 1)^2/4 + 9(y + 1)^2/12 = 1\]into standard form, the terms
- \((x - 1)\) give an \(h\) value of 1,
- and \((y + 1)\) give a \(k\) value of -1.
Semi-major and Semi-minor Axes
The semi-major and semi-minor axes are the backbone of an ellipse's structure, defining its size and shape. In the equation's standard form:\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1,\]the terms \(a^2\) and \(b^2\) are crucial.
- When \(a^2\) is less than \(b^2\), \(a\) is the semi-minor axis and represents the shortest radius of the ellipse.
- Conversely, \(b\) becomes the semi-major axis, representing the longest radius.
- Solving gives \(a = 2\) and the semi-minor axis.
- \(b = \sqrt{12}\) indicates the semi-major axis.
Other exercises in this chapter
Problem 48
Find a set of parametric equations to represent the graph of the given rectangular equation using the parameters (a) \(t=x\) and (b) \(t=2-x.\) $$y=\frac{1}{x^{
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Use a graphing utility to graph the polar equation. Describe your viewing window. $$r=e^{\theta}$$
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Find the standard form of the equation of the hyperbola with the given characteristics. Foci: (-1,3),(9,3) asymptotes: \(y=\frac{3}{4} x, y=6-\frac{3}{4} x\)
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