Problem 40 - Precalculus: Functions and Graphs

Question

Identify the graph of the equation as a parabola (with vertical or horizontal axis), circle, ellipse, or hyperbola. $$x^{2}+4 x+4 y^{2}-24 y=-36$$

Step-by-Step Solution

Answer

The graph represents an ellipse.

Step 1: Rearrange the Equation

Start by rearranging the given equation to group the x-terms and y-terms:\[ x^{2} + 4x + 4y^{2} - 24y = -36 \] Rewrite it as follows: \[ (x^{2} + 4x) + (4y^{2} - 24y) = -36 \]

Step 2: Complete the Square for x

To complete the square for the x-terms in the equation, consider the expression \(x^{2} + 4x\).Find \(\left(\frac{4}{2}\right)^2 = 4\) and rewrite the equation:\[ (x^2 + 4x + 4) = (x + 2)^2 \]

Step 3: Complete the Square for y

Now, complete the square for the y-terms \(4y^2 - 24y\).Factor out the 4 first:\[ 4(y^2 - 6y) \]Find \(\left(\frac{-6}{2}\right)^2 = 9\), so:\[ 4(y^2 - 6y + 9) = 4((y - 3)^2) \]

Step 4: Substitute Completed Squares

Substitute the completed squares back into the equation:\[ (x + 2)^2 + 4((y - 3)^2) = -36 + 4 + 36 \]Simplify the right side:\[ (x + 2)^2 + 4(y - 3)^2 = 4 \]

Step 5: Simplify Equation

Divide the entire equation by 4 to simplify it:\[ \frac{(x + 2)^2}{4} + \frac{(y - 3)^2}{1} = 1 \]This is in the standard form of an ellipse equation.

Key Concepts

Ellipse

An ellipse is a stunning yet simple shape that forms when you take a slice through a cone at an angle, not parallel to the base. It's like an elongated circle. In the equation we worked with, the final result is an ellipse. Every ellipse has two key points called foci, and a line called the major axis, which is the longest diameter. The minor axis is perpendicular to the major one and is the shortest diameter. These axes and the foci control its width and height.

• Ellipse is like a squeezed circle, making it wider or taller.
• The foci are inside the ellipse, always along the major axis.
• The sum of distances from any point on the ellipse to the foci is constant.

Identifying an ellipse in an equation form often involves squaring terms, and if each variable is squared and added, yet with different coefficients, you are likely dealing with an ellipse. So, in our simplified problem, dividing the whole equation by 4 aligns it to the standard form where you can easily identify it as an ellipse.

Completing the Square

Completing the square is like solving a puzzle by arranging the terms neatly to identify shapes. When you complete the square, you're rewriting a quadratic in such a way that makes the expression inside a perfect square trinomial.For example, with our equation, we handle the terms with the same variable one at a time:

• For the expression \(x^2 + 4x\), you find a number that forms a perfect square trinomial like \((x+2)^2\).
• The same trick is applied for the y-terms where \(4(y^2 - 6y)\) becomes \(4((y-3)^2)\).

Why do we do this? It simplifies equations and confirms shapes, like turning our messy equation into the neat form of an ellipse. This step makes life easier, allowing equations to be instantly recognized by form.

Standard Form

Do you ever wish there was a neat book-cover kind of way to write equations? That would be the "standard form." For ellipses, the standard form is particularly tidy and easy to compare. Once we have our ellipse equation:\[ \frac{(x + 2)^2}{4} + \frac{(y - 3)^2}{1} = 1 \]It distinctly reveals the basic structure unique to ellipses in equations. The standard form tells you crucial things right away:

• The denominators \(4\) and \(1\) reveal lengths of axes; larger number for the major axis.
• The values -2 and -3 inside the square show shifts along x and y from the origin.
• The constant 1 on the right is always for ellipses.

When presented in standard form, you can easily deduce the geometric nature and critical points of the conic section described. It's like unplugging a mess to reveal a clear image of what's at hand.