Problem 42 - Contemporary Precalculus

Question

Identify the conic section and use technology to graph it. $$9 x^{2}+y^{2}-36 x+10 y+52=0$$

Step-by-Step Solution

Answer

Given equation: $$9 x^{2}+y^{2}-36 x+10 y+52=0$$ Answer: The given equation represents an ellipse with center at (2, -5), a horizontal major axis, a minor axis length of a = 1, and a major axis length of b = 3.

Step 1: Compare given equation to general equation of different conic sections

Given equation: $$9 x^{2}+y^{2}-36 x+10 y+52=0$$ It is clear that the equation contains both \(x^2\) and \(y^2\) terms, which means it represents a circle, ellipse, or hyperbola. The fact that the coefficients of the \(x^2\) and \(y^2\) terms have the same sign and are unequal indicates that it represents an ellipse.

Step 2: Rewrite the equation in standard form

To rewrite the equation in standard form for an ellipse, we'll complete the square for both the \(x\) and \(y\) terms. We can do this by: 1. Grouping the x and y terms: \((9x^2 - 36x) + (y^2 + 10y) = -52\) 2. Complete the square: Add and subtract the square of half the linear coefficient divided by the quadratic coefficient of both x and y. Half the linear coefficients: \(-\frac{36}{2}= -18\) for x and \(\frac{10}{2} = 5\) for y. Divide by quadratic coefficients: \(\frac{-18}{9}= -2\) for x, y remains 5 (coefficient is 1). Squares of the divided linear coefficients: \((-2)^2 = 4\) for x and \(5^2 = 25\) for y. $$\Rightarrow (9x^2 - 36x + 36) + (y^2 + 10y + 25) = -52 + 36 + 25$$ $$\Rightarrow 9(x-2)^2 + (y+5)^2 = 9$$ 3. Divide by the constant on the right side of the equation so that it equals 1 $$\Rightarrow \frac{(x-2)^2}{1} + \frac{(y+5)^2}{9} = 1$$ Now the equation is in the standard form of an ellipse: \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), with \(h = 2\), \(k = -5\), \(a^2 = 1\), and \(b^2 = 9\).

Step 3: Graph the ellipse using technology

Now that we have the equation in standard form, we can use technology to graph the ellipse. You can use an online graphing calculator, such as Desmos, to plot the standard form of the equation: $$\frac{(x-2)^2}{1} + \frac{(y+5)^2}{9} = 1$$ Once the equation is graphed, you'll see an ellipse with center at \((2, -5)\), a horizontal major axis, a minor axis length of \(a = 1\), and major axis length of \(b = 3\).

Key Concepts

Standard Form of an Ellipse

Understanding the standard form of an ellipse is essential for analyzing and graphing this type of conic section. The standard form is an algebraic expression that allows us to easily identify the properties of an ellipse. In general, the standard form of an ellipse is written as:
\begin{align*}\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} &= 1 \end{align*}

where

• \((h, k)\) is the center of the ellipse,
• \(a\) is the semi-major axis length if \(a > b\), otherwise it is the semi-minor axis length,
• \(b\) is the semi-minor axis length if \(a > b\), otherwise it is the semi-major axis length,
• And \(a^2\) and \(b^2\) are squared lengths of the respective axes.

Placing an ellipse’s equation into this form highlights its key characteristics, such as the orientation, center, and lengths of the major and minor axes. Using this form simplifies the process of graphing ellipses and analyzing their geometric properties.

Complete the Square

The technique of completing the square is a powerful algebraic tool used for transforming quadratic equations into a form that reveals additional information about their graphs. To complete the square, you manipulate a quadratic expression of the form:
\begin{align*}ax^2 + bx + c &= 0\end{align*}

into the form:
\begin{align*}a(x - h)^2 + k &= 0\end{align*}
This is done by:

1. Dividing all terms by \(a\) if \(a eq 1\), to make the coefficient of \(x^2\) equal to 1.
2. Add and subtract the square of half of the coefficient of \(x\) within the parenthesis.
3. Group the perfect square trinomial and move the constant term to the other side of the equation.

By completing the square for elliptical equations, you effectively transition from a general quadratic equation to the standard form of an ellipse, setting the stage for graphing and further analysis.

Conic Sections in Precalculus

In precalculus, conic sections are curves obtained by intersecting a plane with a cone. There are four basic types of conic sections: circles, ellipses, parabolas, and hyperbolas. Each has its unique standard form and set of defining properties:

• Circle: \(x^2 + y^2 = r^2\), where \(r\) is the radius.
• Ellipse: \(\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\), where \(a\) and \(b\) are the lengths of the axes.
• Parabola: \(y = ax^2 + bx + c\), for a vertical parabola, or \(x = ay^2 + by + c\), for a horizontal one.
• Hyperbola: \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\), where the transverse and conjugate axes determine the shape.

Recognizing the form of a conic section is critical before moving on to more detailed work such as graphing or finding focus and directrix. Proper understanding of conic sections paves the way for explorations into more complex mathematical and physical concepts.

Ellipse Graphing

Successfully graphing an ellipse involves conveying its size, shape, and location based on its standard form equation. Here are the basic steps to graph an ellipse:

1. Identify and plot the center \((h, k)\).
2. Plot the vertices by moving \(a\) units horizontally from the center if the greater denominator is under the \(x\)-term, or \(b\) units vertically if the opposite is true.
3. Sketch the co-vertices by moving \(b\) units vertically or \(a\) units horizontally, opposite to how you plotted the vertices.
4. Connect the vertices and co-vertices to form an oval shape, using the center as a guide to ensure symmetry.

Software and graphing calculators, like Desmos, can be used to input the standard form of an ellipse and automatically generate a graph. This technology is a helpful validation tool for understanding the results of hand-drawn sketches and offers an interactive means to explore the effects of changing different values in an ellipse’s equation.