Problem 10 - College Algebra

Question

For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. $$ 4 x^{2}-8 x+9 y^{2}-72 y+112=0 $$

Step-by-Step Solution

Answer

Yes, it's an ellipse: \(\frac{(x - 1)^2}{9} + \frac{(y - 4)^2}{4} = 1\).

Step 1: Identify the Structure of the Equation

Observe the equation \(4x^2 - 8x + 9y^2 - 72y + 112 = 0\). It resembles a conic section equation, which includes terms with \(x^2\) and \(y^2\). For ellipses, the coefficients of \(x^2\) and \(y^2\) must be positive, which they are here.

Step 2: Rearrange and Separate Terms

Rearrange the equation by grouping \(x\)-related and \(y\)-related terms together: \(4x^2 - 8x + 9y^2 - 72y = -112\). Your goal is to complete the square for both \(x\) and \(y\) variables.

Step 3: Complete the Square for \(x\) Terms

Focus on \(4x^2 - 8x\) from the equation, and factor out the 4: \(4(x^2 - 2x)\). To complete the square, add and subtract \(1\) inside the parentheses, getting \(4((x - 1)^2 - 1)\), which simplifies to \(4(x - 1)^2 - 4\).

Step 4: Complete the Square for \(y\) Terms

Now focus on \(9y^2 - 72y\) and factor out the 9: \(9(y^2 - 8y)\). Complete the square by adding and subtracting \(16\) within the parentheses, resulting in \(9((y - 4)^2 - 16)\), simplifying to \(9(y - 4)^2 - 144\).

Step 5: Form the Equation

Insert the completed squares back into the equation: \(4(x - 1)^2 - 4 + 9(y - 4)^2 - 144 = -112\). Simplify: \(4(x - 1)^2 + 9(y - 4)^2 = 36\).

Step 6: Divide by 36 to Obtain Standard Form

Divide the entire equation by 36 to get the ellipse's standard form: \(\frac{(x - 1)^2}{9} + \frac{(y - 4)^2}{4} = 1\). This confirms an ellipse since the equation matches \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), where both denominators are positive.

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a cone. These sections can create various shapes, including circles, ellipses, parabolas, and hyperbolas. The equation provided in the exercise, \[ 4x^2 - 8x + 9y^2 - 72y + 112 = 0 \] initially resembles a conic section as it includes squared terms, specifically terms in both \(x^2\) and \(y^2\).- For a conic to be an ellipse, the coefficients of \(x^2\) and \(y^2\) must both be positive. In this particular problem, the conic section is an ellipse since both coefficients are indeed positive. Understanding this is key to determining which form the equation ultimately represents.

Completing the Square

Completing the square is an algebraic technique used to transform quadratic equations into a perfect square trinomial. This method plays a crucial role in rewriting the equation into the standard form of an ellipse.To complete the square for the equation, you must perform it separately for the \(x\) and \(y\) terms:

• For \(x\): Take the terms \(4x^2 - 8x\). Factor out the 4 to simplify to \(4(x^2 - 2x)\).
• Add and subtract 1 inside the parentheses to complete the square, yielding \(4((x - 1)^2 - 1)\).

For the \(y\) terms:

• For \(y\): Consider \(9y^2 - 72y\) and factor out the 9, resulting in \(9(y^2 - 8y)\).
• To complete the square, add and subtract 16 within, giving \(9((y - 4)^2 - 16)\).

Completing the square is essential for simplifying and eventually converting the conic equation into its standard form.

Standard Form of Ellipse

The standard form of an ellipse is given by\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\]where \((h,k)\) is the center of the ellipse, and \(a\) and \(b\) are the semi-major and semi-minor axes.After completing the square for both \(x\) and \(y\) terms and simplifying the problem's equation, we rewrite it to:\[\frac{(x - 1)^2}{9} + \frac{(y - 4)^2}{4} = 1.\]In this equation:

• The center of the ellipse is at \((1, 4)\).
• The semi-major axis is along the \(x\)-direction with length 3 (since \(\sqrt{9} = 3\)).
• The semi-minor axis is along the \(y\)-direction with length 2 (since \(\sqrt{4} = 2\)).

Knowing the standard form helps in visualizing the ellipse's orientation and dimensions immediately.

Algebraic Manipulation

Algebraic manipulation involves strategically rewriting expressions to make them more readily solvable or interpretable. This exercise requires considerable algebraic manipulation to transform the original conic section equation into the standard ellipse form.The operations included:

• Rearranging terms: Grouping like terms in both variables to facilitate completing the square.
• Simplification: Converting completed square forms to reduce them from additions and subtractions back to a sum equated to -112.
• Balancing the equation: Adjust terms so both sides remain equivalent after new terms from the completion process are introduced.
• Division: Finally, dividing through by the constant (36 here) to standardize the equation to the classic ellipse form, matching it closely to \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\).

Algebraic manipulation is a powerful tool in shifting equations into forms that reveal deeper algebraic properties and geometric displays.