Problem 18 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ x^{2}+6 x+9 y^{2}-36 y+36=0 $$

Step-by-Step Solution

Answer

Vertices: (0, 2), (-6, 2); Minor axis endpoints: (-3, 3), (-3, 1); Foci: (-3 ± 2√2, 2).

Step 1: Rewrite the equation in standard form

First, rearrange the given equation to complete the square. The given ellipse is: \[ x^2 + 6x + 9y^2 - 36y + 36 = 0 \]Group the \(x\) and \(y\) terms separately:\[ (x^2 + 6x) + (9y^2 - 36y) = -36 \]Complete the square for both groups.

Step 2: Complete the square for x-terms

Take the \(x\)-terms:\[ x^2 + 6x \] To complete the square, add and subtract \((\frac{6}{2})^2=9\)\[ x^2 + 6x + 9 - 9 \] This simplifies to:\[ (x + 3)^2 - 9 \]

Step 3: Complete the square for y-terms

Take the \(y\)-terms:\[ 9y^2 - 36y \]Factor out the 9:\[ 9(y^2 - 4y) \] To complete the square inside the parenthesis, add and subtract \((\frac{4}{2})^2 = 4\):\[ 9((y^2 - 4y + 4) - 4) \]This will be:\[ 9((y - 2)^2 - 4) \] Which can be expanded to:\[ 9(y - 2)^2 - 36 \]

Step 4: Substitute and combine terms

Substitute the results from Step 2 and Step 3 into the equation:\[ (x + 3)^2 - 9 + 9(y - 2)^2 - 36 = -36 \]Simplify the equation:\[ (x + 3)^2 + 9(y - 2)^2 = 9 \]

Step 5: Divide by 9 to obtain the standard form

Divide each term by 9 to simplify the equation:\[ \frac{(x + 3)^2}{9} + \frac{(y - 2)^2}{1} = 1 \]This is now in the standard form of an ellipse centered at \((h, k) = (-3, 2)\) with \(a^2 = 9\) and \(b^2 = 1\).

Step 6: Identify and calculate the ellipse parameters

From the standard form, \(\frac{(x + 3)^2}{9} + \frac{(y - 2)^2}{1} = 1\), identify:- Center: \((-3, 2)\)- \(a^2 = 9\) gives \(a = 3\)- \(b^2 = 1\) gives \(b = 1\)- \(c^2 = a^2 - b^2 = 9 - 1 = 8\) so \(c = \sqrt{8} = 2\sqrt{2}\)

Step 7: Find vertices, endpoints of minor axis, and foci

Using the center \((-3, 2)\),- Vertices along the major axis (horizontal): \((-3 \pm 3, 2) = (0, 2), (-6, 2)\)- Endpoints of the minor axis (vertical): \((-3, 2 \pm 1) = (-3, 3), (-3, 1)\)- Foci: \((-3 \pm 2\sqrt{2}, 2)\)

Step 8: Sketch the ellipse

Draw an ellipse centered at \((-3, 2)\) with the calculated vertices, endpoints of the minor axis, and foci. The vertex points are \((0, 2)\) and \((-6, 2)\), while the minor axis endpoints are \((-3, 3)\) and \((-3, 1)\), creating the shape of the ellipse tilted along the x-axis.

Key Concepts

Completing the Square

Completing the square is a key technique used in algebra to turn a quadratic expression into a perfect square trinomial. In the context of an ellipse equation, it's used to rewrite terms so the equation can be easily converted into standard form.
To complete the square for a term like \(x^2 + 6x\), you follow these steps:

• Identify the coefficient of the linear term, in this case, 6.
• Divide this coefficient by 2, giving 3, and then square it, resulting in 9.
• Add and subtract this square within the expression: \(x^2 + 6x + 9 - 9\).
• This can be rewritten as \((x + 3)^2 - 9\).

Completing the square for \(y\)-related terms follows a similar process. Factor out the leading coefficient if necessary (e.g., \(9\) in \(9y^2 - 36y\)), and then complete the square within the parenthesis. This allows us to express the \(y\)-related terms in a neat squared format, paving the way for the standard form of the ellipse.

Standard Form of an Ellipse

The standard form of an ellipse's equation enables you to easily identify its properties, such as its center, axes lengths, and orientation. The standard form is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]where \((h, k)\) is the center of the ellipse, \(a\) is the semi-major axis length, and \(b\) is the semi-minor axis length.
In the exercise's example, after completing the square, we obtained the standard form:\[\frac{(x+3)^2}{9} + \frac{(y-2)^2}{1} = 1\]Here, the values reveal that:

• The center \((h, k)\) is \((-3, 2)\).
• The value \(a^2 = 9\) implies \(a = 3\), indicating the major axis runs horizontally.
• The value \(b^2 = 1\) implies \(b = 1\), showing the minor axis is vertical.

The equation forms a visual reference through which we can draw and understand the ellipse's location and size on a graph.

Center of an Ellipse

The center of an ellipse is the point \((h, k)\) at which the ellipse is symmetrically placed in the coordinate plane. It acts like the anchor for the shape's geometry, from which you measure the lengths of the axes.
Finding the center from the standard form is straightforward. In the expression \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\), \((h, k)\) are simply the values that complete the square brackets for both \(x\) and \(y\).
From the standard form in the given problem:\[\frac{(x+3)^2}{9} + \frac{(y-2)^2}{1} = 1\]The center of the ellipse is noted as \((-3, 2)\). Positioning the ellipse correctly in the graph is crucial for determining other features, such as vertices and foci, easing further analysis of the shape.

Vertices of an Ellipse

The vertices of an ellipse are points located along the major axis where the ellipse reaches its maximum width. They are positioned symmetrically at a distance \(a\) from the center.
For an ellipse with a standard form, the vertices can be directly determined using the values of \(a\) and the center \((h, k)\):

• If the major axis is horizontal: vertices are at \((h \pm a, k)\).
• If the major axis is vertical: vertices are at \((h, k \pm a)\).

In the provided problem, the calculated standard form gives:\[(h, k) = (-3, 2),\ a = 3\]Since the major axis is identified as horizontal:

• Vertices are at \((-3 + 3, 2) = (0, 2)\) and \((-3 - 3, 2) = (-6, 2)\).

Understanding where these points lie helps in drafting the ellipse's shape accurately and grasping its alignment and reach on the graph. This fundamental identification leads to a better visualization of the ellipse's dimensions.