Problem 57 - College Algebra

Question

For the following exercises, determine the value of \(k\) based on the given equation. Given \(2 x^{2}+k x y+12 y^{2}+10 x-16 y+28=0\) find \(k\) for the graph to be an ellipse.

Step-by-Step Solution

Answer

\(-4\sqrt{6} < k < 4\sqrt{6}\) for an ellipse.

Step 1: Identify Equation Form

The given equation is in the form of a general second-degree equation in two variables: \(2x^2 + kxy + 12y^2 + 10x - 16y + 28 = 0\). To determine the conditions for it to represent an ellipse, we must examine the coefficients of \(x^2\), \(xy\), and \(y^2\).

Step 2: Conditions for an Ellipse

For the equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) to represent an ellipse, the condition \(B^2 - 4AC < 0\) must be satisfied. Here, \(A = 2\), \(B = k\), and \(C = 12\).

Step 3: Apply Ellipse Condition

Apply the condition \(B^2 - 4AC < 0\). Substituting the values, we get the inequality \(k^2 - 4 \times 2 \times 12 < 0\). Simplify this to \(k^2 - 96 < 0\).

Step 4: Solve the Inequality

Solve \(k^2 - 96 < 0\) for \(k\). This can be rewritten as \(k^2 < 96\), leading to the solution \(-\sqrt{96} < k < \sqrt{96}\). Calculating further, \(\sqrt{96} = 4\sqrt{6}\), therefore \(-4\sqrt{6} < k < 4\sqrt{6}\).

Step 5: Provide the Solution

The possible values for \(k\) lie within the range \(-4\sqrt{6} < k < 4\sqrt{6}\) for the equation to represent an ellipse.

Key Concepts

General Second-Degree Equation

In algebra, a general second-degree equation in two variables typically takes the form:

• A quadratic expression: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\)
• Includes terms that are squared and linearly combined with each other.

This formula represents a variety of conic sections based on the values of the coefficients \(A\), \(B\), and \(C\). Conic sections include shapes like ellipses, circles, parabolas, and hyperbolas. Understanding these equations is crucial for analyzing these geometric figures.
In our exercise, the equation \(2x^2 + kxy + 12y^2 + 10x - 16y + 28 = 0\) follows this standard form. We identify important numbers: \(A = 2\), \(B = k\), and \(C = 12\). These numbers are key to determining what type of conic section the equation represents.

Conditions for Conic Sections

For each type of conic section represented by a general second-degree equation, specific conditions must be satisfied:

• Circle: \(B = 0\) and \(A = C\).
• Ellipse: Different from a circle where \(B^2 - 4AC < 0\).
• Parabola: Characterized by \(B^2 - 4AC = 0\).
• Hyperbola: Has the condition \(B^2 - 4AC > 0\).

In our context, we're interested in satisfying the conditions for an ellipse. That means we need \(B^2 - 4AC < 0\). With \(B = k\), \(A = 2\), and \(C = 12\) from our original equation, we plug these into the formula to explore acceptable values of \(k\). This confirms the presence of an ellipse when the inequality condition fits.

Inequality Solving in Algebra

Solving inequalities like \(B^2 - 4AC < 0\) frequently focuses on isolating a variable. This step is vital in verifying the shape represented by our conic sections.
Here, the goal is to find values for \(k\) that solve the inequality \(k^2 - 96 < 0\). We simplify it to \(k^2 < 96\), aiming to determine the range of values \(k\) can take.
To solve \(k^2 < 96\), we calculate

• \(-\sqrt{96} < k < \sqrt{96}\).
• Solving for \(\sqrt{96}\), we find it equals \(4\sqrt{6}\).

This provides the range for \(k\): \(-4\sqrt{6} < k < 4\sqrt{6}\). The outcome ensures that for these values, the original equation describes an ellipse, thus completing our understanding of the problem's requirements.