Problem 40 - A Graphical Approach to Precalculus with Limits

Question

Write the equation in standard form for an ellipse centered at ( \(h, k\) ). Identify the center and vertices. $$9 x^{2}-36 x+16 y^{2}-64 y-44=0$$

Step-by-Step Solution

Answer

Center: (2, 2); Vertices: (-2, 2), (6, 2), (2, -1), (2, 5).

Step 1: Reorganize the Equation

First, group the terms with x and y together and move the constant to the other side of the equation. The equation becomes: \[ 9x^2 - 36x + 16y^2 - 64y = 44 \]

Step 2: Complete the Square for x

Identify the terms involving x: \(9x^2 - 36x\). Factor out the 9: \[ 9(x^2 - 4x) \]. Complete the square inside the parentheses. Take half of the coefficient of x, square it, and add and subtract that number inside the parentheses: \[ 9(x^2 - 4x + 4 - 4) \]. Rewrite as: \[ 9((x - 2)^2 - 4) \]. Simplifies to: \[ 9(x - 2)^2 - 36 \].

Step 3: Complete the Square for y

Identify the terms involving y: \(16y^2 - 64y\). Factor out the 16: \[ 16(y^2 - 4y) \]. Complete the square inside the parentheses. Take half of the coefficient of y, square it, and add and subtract that number inside the parentheses: \[ 16(y^2 - 4y + 4 - 4) \]. Rewrite as: \[ 16((y - 2)^2 - 4) \]. Simplifies to: \[ 16(y - 2)^2 - 64 \].

Step 4: Combine and Simplify

Substitute the completed squares back into the equation: \[ 9(x - 2)^2 - 36 + 16(y - 2)^2 - 64 = 44 \]. Simplify the equation: \[ 9(x - 2)^2 + 16(y - 2)^2 = 144 \].

Step 5: Divide to Standard Form

Divide every term by 144 to develop the standard form of the ellipse equation:\[ \frac{(x - 2)^2}{16} + \frac{(y - 2)^2}{9} = 1 \].

Step 6: Identify the Center and Vertices

The center of the ellipse is \((h, k) = (2, 2)\). For the vertices, calculate: - Along the x-axis: \((2 - \sqrt{16}, 2) = (-2, 2)\) and \((2 + \sqrt{16}, 2) = (6, 2)\). - Along the y-axis: \((2, 2 - \sqrt{9}) = (2, -1)\) and \((2, 2 + \sqrt{9}) = (2, 5)\).

Key Concepts

Completing the Square

Completing the square is a crucial technique used to transform a quadratic expression into a perfect square trinomial. For an ellipse, we often encounter terms like \(ax^2 + bx\), which need to be reformulated into a perfect square to reveal the standard form. Here's how you complete the square for a term such as \(9x^2 - 36x\):

• Factor out the coefficient of the squared term, 9 in this case: \(9(x^2 - 4x)\).
• Take half of the linear coefficient, \(-4\), square it, which gives you 4, and add and subtract it inside the parentheses: \(9(x^2 - 4x + 4 - 4)\).
• This becomes \(9((x - 2)^2 - 4))\), simplifying to \(9(x - 2)^2 - 36\).

By completing the square for both \(x\) and \(y\) terms, you can transform the ellipse's equation into a more recognizable and workable form.

Standard Form of an Ellipse

The standard form of an ellipse's equation is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where \((h, k)\) is the center.Completing the square allows us to manipulate a given equation into this format, making it easier to identify properties such as the ellipse's center and vertices.
Once terms are grouped and turned into perfect squares, you can equate the entire expression to 1 by dividing each term by suitable values to simplify the numbers outside the squares.
For example, in the solution provided, dividing by 144 achieved this easily, resulting in the standard form, \[\frac{(x - 2)^2}{16} + \frac{(y - 2)^2}{9} = 1\].This step is essential for recognizing how the ellipse sits on the coordinate plane.

Center and Vertices of an Ellipse

After converting an ellipse equation into its standard form, identifying its main characteristics such as the center and vertices becomes straightforward. The center \((h, k)\) of the ellipse can be directly read from the equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\],which in our example is \((2, 2)\).
To find the vertices:

• Along the x-axis, adjust the \(x\)-coordinate by ±\(a\): The vertices occur at \((h \,\pm\, a, k)\)
• For our example, \(a = \sqrt{16}\) yielding vertices at \((-2, 2)\) and \((6, 2)\).
• Similarly, adjust the \(y\)-coordinate by ±\(b\): The vertices occur at \((h, k \,\pm\, b)\)
• Here, \(b = \sqrt{9}\), so the vertices are \((2, -1)\) and \((2, 5)\).

These steps help illustrate the ellipse's dimensions and orientations in coordinate geometry.

Coordinate Geometry

Coordinate geometry, or analytic geometry, plays a vital role in understanding shapes like ellipses on the Cartesian plane. This branch of mathematics uses algebraic equations to describe and analyze geometric figures.
When working with ellipses:

• Start with an equation involving both \(x\) and \(y\).
• Use techniques like completing the square to manipulate these equations into convenient forms.
• Identify the shape's characteristics, such as center, vertices, axes, and orientation.

This process illustrates how algebra interacts with geometry, allowing us to graph and interpret ellipses easily.
Through understanding these relationships, we can analyze different properties of ellipses and solve related problems intuitively and effectively.