Problem 8 - 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. $$4 x^{2}+5 y^{2}-8 x+30 y+29=0$$

Step-by-Step Solution

Answer

The given equation represents an ellipse, and a suitable viewing window for the graph of the ellipse would be \((-2, 4)\) for the x-axis and \((-5, -1)\) for the y-axis.

Step 1: Find the discriminant of the given conic section equation

The given equation is: $$4 x^{2}+5 y^{2}-8 x+30 y+29=0$$ To find the discriminant, we will use the formula for the discriminant of a conic section: $$D = A^2 - B^2$$ Where A is the coefficient of \(x^2\) and B is the coefficient of \(y^2\). In our case, A = 4 and B = 5, so the discriminant is: $$D = 4^2 - 5^2$$

Step 2: Calculate the discriminant value

Compute the value of the discriminant using the values of A and B: $$D = 16 - 25$$ $$D = -9$$

Step 3: Identify the conic section

Based on the value of the discriminant, we can identify the conic section as follows: - If D > 0, it is a hyperbola - If D = 0, it is a parabola - If D < 0, it is an ellipse In our case, D = -9, which is less than 0. Therefore, the conic section is an ellipse.

Step 4: Complete the square to find the standard form of the ellipse equation

Now, to obtain the standard form of the ellipse equation, we'll complete the square: $$4(x^2 - 2x) + 5(y^2 + 6y) + 29 = 0$$ Next, we complete the square for the x and y terms: $$4[(x - 1)^2 - 1] + 5[(y + 3)^2 - 9] + 29 = 0$$ Now, simplify the equation: $$4(x - 1)^2 - 4 + 5(y + 3)^2 - 45 + 29 = 0$$ $$4(x - 1)^2 + 5(y + 3)^2 = 20$$ Finally, divide by 20 to get the standard form: $$\frac{(x - 1)^2}{5} + \frac{(y + 3)^2}{4} = 1$$

Step 5: Find a suitable viewing window for the graph

Based on the standard form of the ellipse equation, the center of the ellipse is at (1, -3), and the semi-major axis, a, and semi-minor axis, b, are \(\sqrt{5}\) and \(\sqrt{4}\), respectively. To find a suitable viewing window that shows the complete graph, we can set the horizontal and vertical ranges accordingly: Horizontal range: \((-\sqrt{5}+1, \sqrt{5}+1)\) Vertical range: \((-\sqrt{4}-3, \sqrt{4}-3)\) Thus, a suitable viewing window for the graph of the ellipse would be \((-2,4)\) for the x-axis and \((-5,-1)\) for the y-axis.

Key Concepts

Discriminant

The discriminant is a useful mathematical tool that helps in identifying the type of conic section from a given second-degree equation. To find the discriminant of a conic section, we use the formula: \(D = A^2 - B^2\).
Where:

• \(A\) is the coefficient of the \(x^2\) term.
• \(B\) is the coefficient of the \(y^2\) term.

The discriminant tells us:

• If \(D > 0\), the conic section is a hyperbola.
• If \(D = 0\), it is a parabola.
• If \(D < 0\), the conic section is an ellipse.

In the exercise, the discriminant was found to be \(-9\). Since \(D < 0\), this confirms that the conic section is an ellipse.

Ellipse

An ellipse is one of the four types of conic sections and is often described as an elongated circle. In mathematical terms, it consists of all points for which the sum of the distances to two fixed points (the foci) is constant.
Ellipses have a standard equation, which is particularly useful for graphing and analyzing their properties.
Ellipses have the following distinctive features:

• The longest diameter, known as the major axis, determines its width.
• The shortest diameter, the minor axis, determines its height.
• The center is midway between the foci.

In the solution, it was determined that the original equation represents an ellipse by calculating the discriminant. This helps in configuring the graph's display parameters, such as the center and axes lengths.

Standard Form

The standard form of a conic section provides a clear way to recognize and analyze its properties.
For an ellipse, the standard form is expressed as:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here, \((h, k)\) is the center of the ellipse. The values \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.
Converting a given equation to the standard form makes it easier to:

• Identify the ellipse's orientation and dimensions.
• Determine the center, axes lengths, and necessary adjustments for plotting the graph.

In the solution, the equation was rearranged using this process, revealing the center of the ellipse at \((1, -3)\), and axes defined by \(a^2 = 5\) and \(b^2 = 4\).

Completing the Square

Completing the square is a mathematical method used to transform quadratic expressions into a perfect square trinomial. This technique is critical in reaching the standard form of an equation in conic sections.
In conic sections like ellipses, completing the square involves:

• Grouping \(x\) and \(y\) terms respectively.
• Rewriting each quadratic part as a squared binomial.
• Ensuring the equation aligns with the standard form adapted for ellipses.

To achieve this, balance the equation by adjusting constants, so every squared term has a corresponding completive term.
For instance, the step-by-step solution applied this technique:

• For \(x^2 - 2x\): resulting in \((x - 1)^2\).
• For \(y^2 + 6y\): resulting in \((y + 3)^2\).

This approach simplifies the expression to the form \(\frac{(x - 1)^2}{5} + \frac{(y + 3)^2}{4} = 1\), indicating it is ready for graphing.