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Chapter 10: Problem 19

Use the process of completing the square to transform the given equation to a standard form. Then name the corresponding curve and sketch its graph. $$ 4 x^{2}+4 y^{2}-24 x+36 y+81=0 $$

Short Answer

Expert verified
The curve is a circle with center (3, -4.5) and radius 3.

Step by step solution

01

Group variables

First, group the terms involving x and y:\[ 4(x^2 - 6x) + 4(y^2 + 9y) + 81 = 0 \] This separates x-terms and y-terms for easy completion of the square.
02

Move constant to the other side

We need to isolate the constant term from the variables. Move 81 to the right-hand side of the equation:\[ 4(x^2 - 6x) + 4(y^2 + 9y) = -81 \]
03

Complete the square for x-terms

For the expression \( x^2 - 6x \), take half of -6, square it, and add/subtract it inside the equation:\[ x^2 - 6x + 9 - 9 \]This can be rewritten as:\[ (x - 3)^2 - 9 \]
04

Complete the square for y-terms

For \( y^2 + 9y \), take half of 9, square it, and add/subtract:\[ y^2 + 9y + \frac{81}{4} - \frac{81}{4} \]Rewrite this as:\[ \left(y + \frac{9}{2}\right)^2 - \frac{81}{4} \]
05

Simplify and factor

Substitute the completed squares back into the equation:\[ 4((x - 3)^2 - 9) + 4\left(\left(y + \frac{9}{2}\right)^2 - \frac{81}{4}\right) = -81 \] Simplify by factoring and reducing further:\[ 4(x - 3)^2 - 36 + 4\left(y + \frac{9}{2}\right)^2 - 81 = -81 \] \[ 4(x - 3)^2 + 4\left(y + \frac{9}{2}\right)^2 = 36 \] Divide everything by 4:\[ (x - 3)^2 + \left(y + \frac{9}{2}\right)^2 = 9 \]
06

Identify and sketch the curve

The equation \( (x - 3)^2 + \left(y + \frac{9}{2}\right)^2 = 3^2 \) is the equation of a circle with a center at \((3, -\frac{9}{2})\) and a radius of 3.Sketching it would show a circle centered at this point with a radius of 3 units.

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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 technique is especially helpful when dealing with quadratic equations and conic sections, such as circles and parabolas. To complete the square, follow these basic steps for a given quadratic expression:
  • Identify the quadratic and linear coefficients. For example, consider the expression \( x^2 - 6x \). Here, the coefficient of \( x \) (the linear term) is -6.
  • Take half of the linear coefficient (in this case, -6 梅 2 = -3) and square it. This gives \( (-3)^2 = 9 \).
  • Add and subtract this squared result within the expression: \( x^2 - 6x + 9 - 9 \).
  • Factor it into the form \((x - 3)^2 - 9\), where \((x - 3)^2\) is a perfect square trinomial.
This method is repeated for other variables such as \( y \) in our example, making it simpler to recognize and solve equations involving conic sections by revealing their standard forms.
Equation of a Circle
The equation of a circle in standard form is written as \[(x - h)^2 + (y - k)^2 = r^2\]where:
  • \((h, k)\) is the center of the circle.
  • \(r\) is the radius of the circle.
In our specific exercise, we transformed the given equation into the form:\[(x - 3)^2 + \left(y + \frac{9}{2}\right)^2 = 3^2\]This tells us immediately that the circle is centered at \((3, -\frac{9}{2})\) with a radius of 3. This standard form makes it easy to identify both the location of the circle on the coordinate plane and its size. Recognizing this form can also simplify graphing and understanding circles, as the center and radius give a clear visual representation of the circle's properties.
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to achieve a desired form. In the process of finding the standard form of a circle's equation, algebraic skills are crucial to handle terms efficiently.In our example, the steps involved:
  • Grouping similar terms. This helped in separating the \(x\) and \(y\) terms to work on them individually, essential before completing the square.
  • Moving constants to the other side of the equation. For clarity, we moved \(81\) to isolate variable-related terms.
  • Carefully factoring out and rearranging squares. This simplified the algebraic expression, allowing for the identification of perfect squares and the ultimate reduction to the standard circle equation.
Algebraic manipulation is a powerful tool in mathematics that ensures flexibility and depth of understanding. It empowers students to approach equations wisely and make connections between various mathematical concepts.

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Most popular questions from this chapter

Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(x^{2}-4 y^{2}=4\)

Find the coordinates of the focus and the equation of the directrix for each parabola. Make a sketch showing the parabola, its focus, and its directrix. \(y^{2}=x\)

Using the same axes, draw the conics \(y=\) \(\pm\left(a x^{2}+1\right)^{1 / 2}\) for \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\) using \(a=\) \(-2,-1,-0.5,-0.1,0,0.1,0.6,1 .\) Make a conjecture about how the shape of the figure depends on \(a\).

Name the conic (horizontal ellipse, vertical hyperbola, and so on ) corresponding to the given equation. \(9 x^{2}+4 y^{2}=9\)

Find the equations of the tangent and the normal lines to the given parabola at the given point. Sketch the parabola, the tangent line, and the normal line. \(x^{2}=4 y,(4,4)\)
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