Problem 36 - Contemporary Precalculus

Question

Use the information given in Special Topics 10.3. A and summarized in the endpapers at the beginning of this book to find a parameterization of the conic section whose rectangular equation is given. Confirm your answer by graphing. $$x^{2}+y^{2}-4 x-6 y+9=0$$

Step-by-Step Solution

Answer

Answer: The parameterization for the given conic section is $$x(t) = 2 + 2\cos t$$ and $$y(t) = 3 + 2\sin t$$ with $$t\in[0, 2\pi]$$. To confirm this, graph the original equation and the parameterized equations in the same coordinate system. They represent the same circle with center (2,3) and radius 2, confirming the correctness of our parameterization.

Step 1: Completing the square

To complete the squares for the x and y portions of the equation, rearrange and group terms for each variable: $$x^2 - 4x + y^2 - 6y = -9$$ Now, complete the square for each variable: $$\left(x - 2\right)^2 - 4 + \left(y - 3\right)^2 - 9 = -9$$ Then, simplify the equation: $$ \left(x - 2\right)^2 + \left(y - 3\right)^2 = 4$$

Step 2: Identify the type of conic section

We have an equation of the form: $$\left(x - 2\right)^2 + \left(y - 3\right)^2 = 4$$ This is the equation of a circle with center at (2, 3) and radius 2.

Step 3: Find the parameterization

We will use the general parameterization for a circle, given by: $$x = x_0 + r * \cos t$$ $$y = y_0 + r * \sin t$$ Plug in our center (2,3) and radius (2) to get the parameterization for the circle: $$x = 2 + 2 * \cos t$$ $$y = 3 + 2 * \sin t$$ So, our parameterization for the given conic section is: $$x(t) = 2 + 2\cos t$$ $$y(t) = 3 + 2\sin t$$ with $$t\in[0, 2\pi]$$

Step 4: Confirm through graphing

To confirm our answer, we must graph the original equation and the parameterized equation to check if they represent the same circle. Graph $$x^2 + y^2 - 4x - 6y + 9 = 0$$ and the parameterized equations $$x(t) = 2 + 2\cos t$$ and $$y(t) = 3 + 2\sin t$$ in the same coordinate system. When you graph both equations, you will see that they represent the same circle with center (2,3) and radius 2, confirming the correctness of our parameterization.

Key Concepts

Completing the Square

Completing the square is a powerful technique used to transform quadratic equations, making them easier to understand and solve. When you see a quadratic equation involving terms like \(x^2\) or \(y^2\), completing the square helps in identifying the nature of the conic section the equation represents.

For the given equation \(x^2 + y^2 - 4x - 6y + 9 = 0\), we follow these steps:

• Group the \(x\) and \(y\) terms separately: \(x^2 - 4x\) and \(y^2 - 6y\).
• To complete the square for each group, adjust the equation by adding and subtracting the necessary constants within each group. For \(x^2 - 4x\), add and then subtract 4 to form \((x - 2)^2 - 4\); for \(y^2 - 6y\), add and then subtract 9 to form \((y - 3)^2 - 9\).

This conversion transforms the equation into the standard circle form \((x - 2)^2 + (y - 3)^2 = 4\), revealing that the equation represents a circle. Completing the square distinguishes between different conic sections, like circles, ellipses, parabolas, and hyperbolas.

Parameterization

Parameterization is a method used to express mathematical objects, like curves, using a parameter. For a circle, this process highlights its continuous nature and position in a coordinate system.

Once we identified that our equation \((x - 2)^2 + (y - 3)^2 = 4\) represents a circle, we can parameterize it. **Parameterization of a Circle** involves:

• Using a parameter \(t\) that typically ranges between \(0\) and \(2\pi\).
• Representing \(x\) and \(y\) using trigonometric functions: \(x(t) = x_0 + r \cos t\), \(y(t) = y_0 + r \sin t\), where \(x_0\) and \(y_0\) are the center coordinates and \(r\) is the radius.

For our circle with center \((2, 3)\) and radius \(2\), the parameterization becomes:
\[ x(t) = 2 + 2\cos t \]
\[ y(t) = 3 + 2\sin t \]
This offers a clear way to plot the circle or compute various properties by simply adjusting \(t\).

Circle Equation

A circle equation in standard form reveals both the center and radius, providing a comprehensive view of a circle's geometry. In algebraic terms, the standard form is \((x - x_0)^2 + (y - y_0)^2 = r^2\), where \((x_0, y_0)\) is the center and \(r\) is the radius.

In our exercise, after completing the square transformed \(x^2 + y^2 - 4x - 6y + 9 = 0\) into \((x - 2)^2 + (y - 3)^2 = 4\), we observed:

• The center at \((2, 3)\).
• The radius \(2\), since \(4\) is the square of \(r\).

This highlights how processing the equation provides us direct insight into the circle's geometry. Such equations are extensively used in geometry, design, and various applications, offering a predictable and consistent way to handle circular shapes.