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Chapter 11: Problem 64

Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections. \(4 x^{2}+y^{2}+8 x-2 y=-1\)

Short Answer

Expert verified
The center is (-1, 1), vertices are \((-1\pm\frac{\sqrt{5}}{2}, 1)\) and \((-1, 1\pm\sqrt{5})\), and foci are at \((-1\pm\sqrt{\frac{15}{4}}, 1)\).

Step by step solution

01

Identify the Conic Section

To identify the type of conic section, we start by examining the given equation \(4x^{2} + y^{2} + 8x - 2y = -1\). We notice that there is both a \(x^2\) and a \(y^2\) term, suggesting that this is either an ellipse or a hyperbola. The coefficients of \(x^2\) and \(y^2\) are not equal, suggesting that this is an ellipse.
02

Rearrange the Equation

Rearrange the given equation to make completing the square possible. Start by getting all terms involving \(x\) and \(y\) on one side of the equation: \(4x^2 + 8x + y^2 - 2y = -1\). Then, move the constant to the other side: \(4x^2 + 8x + y^2 - 2y + 1 = 0\).
03

Complete The Square

For the terms involving \(x\): Factor out the \(4\), giving \(4(x^2 + 2x)\). Complete the square: \((x + 1)^2 = x^2 + 2x + 1\). This adds 4 to the equation: \(4((x+1)^2) - 4\). For the terms involving \(y\): Take \(y^2 - 2y\) and complete the square: \((y - 1)^2 = y^2 - 2y + 1\). Add and subtract 1 inside the square, resulting in \((y-1)^2 - 1\).
04

Rewrite the Equation

After completing the square, rewrite the equation: \[4((x+1)^2 - 1) + ((y-1)^2 - 1) = -1\]\[4(x+1)^2 - 4 + (y-1)^2 - 1 = -1\]Combine: \[4(x+1)^2 + (y-1)^2 = 5\].This equation can be rewritten in standard form as: \[\frac{(x+1)^2}{\frac{5}{4}} + \frac{(y-1)^2}{5} = 1\].
05

Identify Key Features

From the equation \(\frac{(x+1)^2}{\frac{5}{4}} + \frac{(y-1)^2}{5} = 1\), we find:- **Center**: (-1, 1)- The terms under the squares in the equation represent the squares of the semi-axes.- **Vertices**: The square root of \(\frac{5}{4}\) is \(\frac{\sqrt{5}}{2}\) and the square root of \(5\) is \(\sqrt{5}\). So, vertices along the x-axis at \((-1 \pm \frac{\sqrt{5}}{2}, 1)\) and along the y-axis at \((-1, 1 \pm \sqrt{5})\).- **Foci**: Calculated using \(c^2 = a^2 - b^2\), where \(a = \sqrt{5}\) and \(b = \frac{\sqrt{5}}{2}\). This results in \(c = \sqrt{\frac{15}{4}}\), giving foci at \((-1 \pm\sqrt{\frac{15}{4}}, 1)\).- **Asymptotes**: Not applicable for ellipses.- **Radius**: Since this is an ellipse, the concept of radius is replaced by semi-major and semi-minor axes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conic Sections
Conic sections are shapes created by cutting through a cone with a plane. There are four basic types to consider:
  • Circle: A special type of ellipse with equal axes.
  • Ellipse: An elongated circle with two different axes lengths.
  • Parabola: A mirror-symmetrical, open curve.
  • Hyperbola: Comprised of two separate curves facing away from each other.
The equation given in our exercise is initially not in any recognizable standard form. By observing both squared terms, it suggests an ellipse or hyperbola. Since the coefficients for the squared terms are different, we identify it as an ellipse. Ellipses are defined by their center, vertices, and foci, emphasizing their elongated, oval shape. Our task is to transform this equation to reveal these characteristics.
Completing the Square
Completing the square is a method used to transform quadratic equations into a form that makes it easy to identify conic sections, like ellipses in this case. It involves making a perfect square trinomial from a standard quadratic expression. We start by rearranging terms to isolate the variables on one side. For example, the given equation, after being rearranged, becomes \(4x^2 + 8x + y^2 - 2y = -1\). Next, each part involving a variable is turned into a perfect square:
  • For terms with \(x\): Factor out any constant from the squared term (here, 4), yielding \(4(x^2 + 2x)\). Then, complete the square: \((x+1)^2 = x^2 + 2x + 1\), which affects the equation's balance.
  • For terms with \(y\): Similarly, transform \(y^2 - 2y\) into \((y-1)^2 = y^2 - 2y + 1\).
This transformation allows the equation to reveal its core features, expressing it in a standard form of an ellipse: \(\frac{(x+1)^2}{5/4} + \frac{(y-1)^2}{5} = 1\). Completing the square elegantly simplifies the understanding of this conic section.
Vertices of an Ellipse
Vertices in an ellipse are points where the ellipse is widest along one direction. In simpler terms, they are like the tips of an elongated oval. For a standard ellipse equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\):
  • The vertices along the major axis, determined by the larger denominator, are given by \((h \pm a, k)\) or \((h, k \pm b)\), depending on the position of \(a\) and \(b\).
In the provided problem, after identifying the equation in its standard form, the vertices can be calculated:
  • Major axis along the y-direction: \((-1, 1 \pm \sqrt{5})\)
  • Minor axis along the x-direction: \((-1 \pm \frac{\sqrt{5}}{2}, 1)\)
Vertices are crucial because they symbolize the maximum stretch of the ellipse, making them essential in understanding its shape and orientation.
Foci of an Ellipse
The foci of an ellipse are two internal points used to define and construct the ellipse. The defining property of an ellipse is that the sum of the distances from any point on the ellipse to the two foci is constant. To find the foci location:
  • The formula \(c^2 = a^2 - b^2\) helps calculate the distance \(c\) from the center to each focus.
  • For our calculated ellipse, \(a\), the semi-major axis, is \(\sqrt{5}\), and \(b\), the semi-minor axis, is \(\frac{\sqrt{5}}{2}\).
  • Plugging into the formula gives \(c = \sqrt{\frac{15}{4}}\).
  • Thus, the foci positions are at \((-1 \pm \sqrt{\frac{15}{4}}, 1)\).
Understanding the foci helps us get a deeper insight into the elongation and overall structure of the ellipse. They are critical for precise applications in geometry, physics, and engineering.

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Most popular questions from this chapter

Give equations for hyperbolas and tell how many units up or down and to the right or left each hyperbola is to be shifted. Find an equation for the new hyperbola, and find the new center, foci, vertices, and asymptotes. \(y^{2}-x^{2}=1,\) left 1, down 1

A cone frustum The line segment joining the points \((0,1)\) and \((2,2)\) is revolved about the \(x\) -axis to generate a frustum of a cone. Find the surface area of the frustum using the parametrization \(x=2 t, y=t+1,0 \leq t \leq 1 .\) Check your result with the geometry formula: Area \(=\pi\left(r_{1}+r_{2}\right)(\) slant height).

In Exercises \(51-54,\) use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(11.16 . )\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$ x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1 $$

Give information about the foci, vertices, and asymptotes of hyperbolas centered at the origin of the \(x y\)-plane. In each case, find the hyperbola's standard-form equation from the information given. Vertices: \((0, \pm 2)\) Asymptotes: \(y=\pm \frac{1}{2} x\)

The curve with parametric equations $$x=t, \quad y=1-\cos t, \quad 0 \leq t \leq 2 \pi$$ is called a sinusoid and is shown in the accompanying figure. Find the point \((x, y)\) where the slope of the tangent line is a. largest. \(\quad\) b. smallest.
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