Problem 12 - Precalculus: Functions and Graphs

Question

Exer. 1-14: Find the vertices and foci of the ellipse. Sketch its graph, showing the foci. $$x^{2}+2 y^{2}+2 x-20 y+43=0$$

Step-by-Step Solution

Answer

Vertices: \((-1 \pm 2\sqrt{2}, 5)\) and \((-1, 5 \pm 2)\); Foci: \((1, 5)\), \((-3, 5)\).

Step 1: Rearrange the equation

Start with the equation \(x^{2} + 2y^{2} + 2x - 20y + 43 = 0\). Rearrange it by grouping the x's and y's: \((x^2 + 2x) + (2y^2 - 20y) = -43\).

Step 2: Complete the square for x

For the x terms, complete the square: \(x^2 + 2x\) becomes \((x+1)^2 - 1\). Thus, \((x^2 + 2x)\) turns into \((x+1)^2 - 1\).

Step 3: Complete the square for y

Factor out the 2 from the y terms: \(2(y^2 - 10y)\). Complete the square inside the parenthesis: \(y^2 - 10y\) becomes \((y-5)^2 - 25\). Thus, \(2(y^2 - 10y)\) turns into \(2((y-5)^2 - 25) = 2(y-5)^2 - 50\).

Step 4: Simplify the equation

Substitute the completed squares back into the equation: \((x+1)^2 - 1 + 2(y-5)^2 - 50 = -43\). Simplify to get \((x+1)^2 + 2(y-5)^2 = 8\).

Step 5: Express the equation in standard ellipse form

Divide the entire equation by 8 to get it in standard form: \(\frac{(x+1)^2}{8} + \frac{(y-5)^2}{4} = 1\). Identify \(a^2 = 8\) and \(b^2 = 4\), where a represents the semi-major axis.

Step 6: Determine vertices and foci

The center of the ellipse is \((-1, 5)\). The vertices are found using \(\pm a\) and \(\pm b\): Vertices along the x-axis: \((-1 \pm \sqrt{8}, 5)\) or \((-1 \pm 2\sqrt{2}, 5)\), and along the y-axis: \((-1, 5 \pm 2)\). The foci are calculated using \(c = \sqrt{a^2 - b^2} = \sqrt{4} = 2\), resulting in foci \((-1 \pm 2, 5)\).

Step 7: Sketch the ellipse

Draw the ellipse centered at \((-1, 5)\) with vertices calculated in Step 6. Be sure to plot and label the foci at \((1, 5)\) and \((-3, 5)\) for visual clarity.

Key Concepts

Vertices of an Ellipse

Understanding the vertices of an ellipse is crucial when analyzing its geometric properties. The vertices are essentially the points where the ellipse is the widest along either the horizontal or vertical axis.
In an ellipse, you often have two pairs of vertices:

• Major Axis Vertices: These are located at the maximum width of the ellipse.
• Minor Axis Vertices: These occur at the shortest width of the ellipse.

For the ellipse represented by \((x+1)^{2}/8 + (y-5)^{2}/4=1\), the center is at \((-1, 5)\). Typical calculation involves moving from this center by \(\pm \sqrt{a^2}\) and \(\pm \sqrt{b^2}\).

• Here, sideways (horizontal axis) vertices are found: \((-1 \pm 2\sqrt{2}, 5)\)
• While the vertical movement (y-axis) gives: \((-1, 5 \pm 2)\)

Recognizing where these vertices lie helps in confirming the shape's reach and verifying calculations.

Foci of an Ellipse

The foci (singular: focus) of an ellipse are two fixed points located along the major axis. The sum of the distances from any point on the ellipse to each focus remains constant, which is a defining property of an ellipse.
In our exercise, we found the standard form of the ellipse as \(rac{(x+1)^2}{8} + rac{(y-5)^2}{4} = 1\). From here, our task is to determine the foci’s location:

• Using the equation, \(c = \sqrt{a^2 - b^2}\), we determine \(c\).
• Given that \(a^2 = 8\) and \(b^2 = 4\), \(c = \sqrt{4} = 2\).
• The foci are positioned at \((-1 \pm 2, 5)\) along the major axis.

Plotting these points aids in understanding the ellipse's orientation and helps in accurately sketching it on a graph.

Completing the Square

Completing the square is a vital technique used to convert an equation into a more workable form, especially useful in determining the standard form of conics like ellipses.
In the given problem, this method was key to transforming the initial daunting equation \(x^{2}+2y^{2}+2x-20y+43=0\) into something manageable.

• For the \(x\) terms, rearrange and transform: \(x^2 + 2x\) becomes \((x+1)^2 - 1\).
• For the \(y\) terms, first factor out 2: \(2(y^2 - 10y)\), completing it results in \((y-5)^2 - 25\), then scaling leads to \((y-5)^2 - 50\).
• Substitute these completed squares back into the original equation for simplification.

This rearrangement not only simplifies solving the equation but also places it in a form where its geometric properties become apparent.

Standard Form of an Ellipse

The standard form of an ellipse makes it easier to identify and visualize its specific attributes, such as the center, axes, and orientation. The general template is either \((x-h)^2/a^2 + (y-k)^2/b^2 = 1\) or the inverse, depending on whether the major axis is horizontal or vertical.
For our specific ellipse, it was brought to \(rac{(x+1)^2}{8} + rac{(y-5)^2}{4} = 1\).

• The center of the ellipse is given by the point \((-h, k)\), here \((-1, 5)\).
• The denominators, \(a^2\) and \(b^2\), explicitly describe the lengths of the axes; here \(a=\sqrt{8}\) indicates a larger stretch in the horizontal axis.

Understanding the standard form is not just about notation—it's about quickly accessing the ellipse’s dimensions and position.