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Chapter 10: Problem 23

Sketch the graph of the given equation. $$ x^{2}+4 y^{2}-2 x+16 y+1=0 $$

Short Answer

Expert verified
The graph is an ellipse centered at (1, -2) with a semi-major axis 4 along the x-axis and semi-minor axis 2 along the y-axis.

Step by step solution

01

Rewrite the Equation

The given equation is \( x^2 + 4y^2 - 2x + 16y + 1 = 0 \). Start by rewriting it to group the \( x \) and \( y \) terms together: \[ x^2 - 2x + 4(y^2 + 4y) + 1 = 0 \]
02

Complete the Square for x

Focus on the \( x \) terms \( x^2 - 2x \). To complete the square, add and subtract \( (-1)^2 \). This results in:\[ (x-1)^2 - 1 \] Replace \( x^2 - 2x \) with this expression in the equation.
03

Complete the Square for y

For the \( y \) terms, look at \( y^2 + 4y \). To complete the square, you add and subtract \( "+4" \) (which is \( (4/2)^2 \)). Thus,\[ y^2 + 4y = (y+2)^2 - 4 \]Substitute this back into the main equation.
04

Substitute and Simplify the Equation

Substitute the completed squares back into the equation:\[ (x-1)^2 - 1 + 4((y+2)^2 - 4) + 1 = 0 \]This simplifies to:\[ (x-1)^2 + 4(y+2)^2 = 16 \]
05

Identify the Conic Section

The equation \( (x-1)^2 + 4(y+2)^2 = 16 \) is in the form of a standard ellipse equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Identifying:- Center \((h, k) = (1, -2)\)- \( a^2 = 16 \), so \( a = 4 \)- \( b^2 = 4 \), so \( b = 2 \)Thus, this is an ellipse centered at \((1, -2)\) with semi-major axis 4 (along x-axis) and semi-minor axis 2 (along y-axis).
06

Sketch the Graph

To sketch the graph of the ellipse:1. Plot the center of the ellipse at \((1, -2)\).2. Draw the major axis along the x-axis, 4 units right and left of the center.3. Draw the minor axis along the y-axis, 2 units up and down from the center.4. Sketch the ellipse shape fitting these dimensions.

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Key Concepts

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

Completing the Square
Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This helps in simplifying equations, especially when dealing with conic sections like ellipses.

To complete the square for an expression like \(x^2 - 2x\), follow these steps:
  • Take the coefficient of \(x\), which is \(-2\), and divide it by 2, giving you \(-1\).
  • Square the result, so \((-1)^2 = 1\).
  • Add and subtract this square inside the expression: \(x^2 - 2x + 1 - 1 = (x-1)^2 - 1\).
Similarly, for \(y^2 + 4y\), you:
  • Take \(4\), divide by 2 to get \(2\), then square it to obtain \(4\).
  • Add and subtract this from the equation: \(y^2 + 4y + 4 - 4 = (y+2)^2 - 4\).
Completing the square effectively rewrites the quadratic terms into squared terms, making it easier to identify and work with the equation of a conic section.
Conic Sections
Conic sections are curves generated by the intersection of a plane with a cone. The four types of conic sections are circles, ellipses, parabolas, and hyperbolas, each with distinct equations and properties.

Ellipses arise when the plane intersects the cone at an angle, causing a closed, oval-shaped curve. These shapes follow a standard form equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where:
  • \((h, k)\) is the center of the ellipse.
  • \(a\) and \(b\) represent the lengths of the semi-major and semi-minor axes, respectively.
  • If \(a > b\), the ellipse is stretched along the x-axis, and if \(b > a\), it is stretched along the y-axis.
Understanding conic sections through their standard form simplifies the process of analyzing and sketching these curves on a graph.
Ellipse Equation
The ellipse equation is an essential element in describing the shape and features of an ellipse. The equation is typically expressed as \((x-1)^2 + 4(y+2)^2 = 16\).

To rewrite this ellipse equation in standard form, divide each term by 16:
  • \(\frac{(x-1)^2}{16} + \frac{4(y+2)^2}{16} = 1\).
  • Simplifying, this becomes \(\frac{(x-1)^2}{16} + \frac{(y+2)^2}{4} = 1\).
From this equation, it's easy to identify the ellipse's characteristics:
  • Center at \((1, -2)\).
  • \(a^2 = 16\), functioning as the semi-major axis \(a = 4\), aligned along the x-axis.
  • \(b^2 = 4\), functioning as the semi-minor axis \(b = 2\), aligned along the y-axis.
The equation provides a precise description of how the ellipse is oriented and scaled within the coordinate plane.
Graph Sketching
Graph sketching involves plotting the visual representation of an equation on a coordinate plane. For an ellipse, it means identifying key components and then sketching them accordingly.

Begin sketching the graph of an ellipse with these steps:
  • Determine and plot the center, which in this case is \((1, -2)\).
  • From the center, move \(a = 4\) units along the x-axis for the vertices of the major axis, placing them at \((-3, -2)\) and \((5, -2)\).
  • Do the same for the y-axis using \(b = 2\), placing edges at \((1, 0)\) and \((1, -4)\).
  • Connect these points in a smooth, oval shape to produce the complete sketch of the ellipse.
Graph sketching is a vital skill that transforms equations into visual understanding, clearly illustrating the size, orientation, and position of ellipses and other conic sections.

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Most popular questions from this chapter

Consider the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\). (a) Show that its perimeter is $$ P=4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t $$ where \(e\) is the eccentricity. C (b) The integral in part (a) is called an elliptic integral. It has been studied at great length, and it is known that the integrand does not have an elementary antiderivative, so we must turn to approximate methods to evaluate \(P\). Do so when \(a=1\) and \(e=\frac{1}{4}\) using the Parabolic Rule with \(n=4\). (Your answer should be near \(2 \pi\). Why?) (c) Repeat part (b) using \(n=20\).

The path of a projectile fired from level ground with a speed of \(v_{0}\) feet per second at an angle \(\alpha\) with the ground is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=-16 t^{2}+\left(v_{0} \sin \alpha\right) t $$ (a) Show that the path is a parabola. (b) Find the time of flight. (c) Show that the range (horizontal distance traveled) is \(\left(v_{0}^{2} / 32\right) \sin 2 \alpha\). (d) For a given \(v_{0}\), what value of \(\alpha\) gives the largest possible range?

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=4 \sqrt{t}, y=t^{2}+\frac{1}{2 t}, \frac{1}{4} \leq t \leq 1 $$

Plot the points whose polar coordinates are \((3,2 \pi)\), \(\left(2, \frac{1}{2} \pi\right), \quad\left(4,-\frac{1}{3} \pi\right), \quad(0,0),(1,54 \pi), \quad\left(3,-\frac{1}{6} \pi\right), \quad\left(1, \frac{1}{2} \pi\right), \quad\) and \(\left(3,-\frac{3}{2} \pi\right)\).

Show that the polar equation of the circle with center \((c, \alpha)\) and radius \(a\) is \(r^{2}+c^{2}-2 r c \cos (\theta-\alpha)=a^{2}\).
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