Problem 7 - Contemporary Precalculus

Question

Use the discriminant to identify the conic section whose equation is given, and find a viewing window that shows a complete graph. $$9 x^{2}+4 y^{2}+54 x-8 y+49=0$$

Step-by-Step Solution

Answer

The given equation represents an ellipse. An appropriate viewing window for the graph is [-7, 1] for x-values and [-2, 4] for y-values.

Step 1: Identify the values of A, B, and C.

The given equation is \(9x^2 + 4y^2 + 54x - 8y + 49 = 0\). Comparing this with the general form of a conic equation, we have: - A = 9 - B = 0 (there's no xy term) - C = 4

Step 2: Calculate the discriminant.

The discriminant is given as \(\Delta = B^2 - 4AC\). Using the values A = 9, B = 0, and C = 4, the discriminant is: $$\Delta = 0^2 - 4 \times 9 \times 4 = 0 - 144 = -144$$

Step 3: Determine the type of conic section.

Using the value of the discriminant, we can determine the type of conic section as follows: - \(\Delta > 0\): hyperbola - \(\Delta = 0\): parabola - \(\Delta < 0\): ellipse or circle In our case, \(\Delta = -144 < 0\), so the conic section is an ellipse or a circle.

Step 4: Distinguish between an ellipse and a circle.

We can now check if the given conic section is an ellipse or a circle by comparing A and C. If A = C, the equation represents a circle. If A ≠ C, the equation represents an ellipse. In our case, A = 9 and C = 4, so A ≠ C, thus the conic section is an ellipse.

Step 5: Find a viewing window.

To find an appropriate viewing window for the graph of the ellipse, we need to first rewrite the equation in the standard form: $$\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$$ In this case, \(9x^2 + 4y^2 + 54x - 8y + 49 = 0\) can be rewritten as \(\frac{(x+3)^2}{16} + \frac{(y-1)^2}{9} = 1\). From this form, we have: - Center: (-3,1) - Semi-major axis: a = 4 - Semi-minor axis: b = 3 So, an appropriate viewing window for the ellipse should be: - x-min: -3 - 4 = -7 - x-max: -3 + 4 = 1 - y-min: 1 - 3 = -2 - y-max: 1 + 3 = 4 Hence, the viewing window for the graph of the ellipse is [-7, 1] for x-values and [-2, 4] for y-values.

Key Concepts

The Discriminant in Math

One of the effective methods to identify the type of conic section from a quadratic equation is through using the discriminant. The discriminant helps determine if a conic is a circle, an ellipse, a parabola, or a hyperbola. The discriminant for a quadratic equation of the form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) is calculated using the formula: \[ \Delta = B^2 - 4AC \] Here's how the discriminant helps identify different conic sections:

• If \( \Delta > 0 \), the equation represents a hyperbola.
• If \( \Delta = 0 \), it indicates a parabola.
• If \( \Delta < 0 \), the conic is either an ellipse or a circle.

In this exercise, the discriminant calculated as \( \Delta = -144 \) indicates an ellipse, since it’s less than zero. Always remember that the discriminant gives us a quick way to understand the nature of the curve without needing to graph it first.

Ellipse Identification

Once the discriminant confirms that the conic section is an ellipse, the next step is to differentiate it from a circle. While both are identified by a negative discriminant, the distinguishing factor lies in comparing the values of \( A \) and \( C \) from the equation.

• If \( A = C \), the conic section is a circle because both the x and y-axis have the same coefficient, indicating symmetry around the center.
• If \( A eq C \), it's an ellipse due to the difference in coefficients, which impacts the curve's symmetry along the axes.

In our solution, we found \( A = 9 \) and \( C = 4 \). Since they are not equal, the conic section is confirmed as an ellipse. Identifying ellipses accurately is crucial, especially in fields like astronomy and physics where they model planetary orbits and optical properties.

Graphing Conic Sections

Graphing conic sections, such as ellipses, requires knowledge of their geometric properties and equation forms. An ellipse's general equation can be rewritten into a standard form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] Where \((h, k)\) is the center of the ellipse, \(a\) represents the semi-major axis, and \(b\) represents the semi-minor axis. Through completing the square and rearranging terms, you transform the original quadratic equation into this understandable form. Once in standard form, graphing becomes straightforward:

• Identify the center coordinates \((h, k)\).
• Use the lengths of \(a\) and \(b\) to plot the axes.
• Draw the ellipse ensuring it’s aligned with the center and the specified radii.

In the exercise, the equation was rewritten as \(\frac{(x+3)^2}{16} + \frac{(y-1)^2}{9} = 1\). It reveals a center at \((-3, 1)\), a semi-major axis \(a = 4\), and a semi-minor axis \(b = 3\). To properly view and graph this ellipse, choose a viewing window that fully captures its extent, such as \([-7, 1]\) for x-values and \([-2, 4]\) for y-values. Understanding the graphing approach aids in visualizing solutions effectively, providing clarity to mathematical concepts.