Problem 40 - College Algebra

Question

For the following exercises, find the foci for the given ellipses. $$ x^{2}-8 x+25 y^{2}-100 y+91=0 $$

Step-by-Step Solution

Answer

The foci are at \((4 + 2\sqrt{6}, 2)\) and \((4 - 2\sqrt{6}, 2)\)."

Step 1: Rearrange and Complete the Square for X

The given equation for the ellipse is \(x^2 - 8x + 25y^2 - 100y + 91 = 0\). First, we rearrange the equation around the \(x\) terms: \(x^2 - 8x\). To complete the square, take half the coefficient of \(x\), which is \(-8\), divide by 2 to get \(-4\), and square it to get \(16\). Add and subtract 16 within the equation to maintain equality:\[x^2 - 8x + 16 - 16 + 25y^2 - 100y + 91 = 0\].

Step 2: Rearrange and Complete the Square for Y

Next, handle the \(y\) terms: \(25y^2 - 100y\). Factor out the coefficient 25: \(25(y^2 - 4y)\). Take half of \(-4\), square it, and adjust within the bracket, remember to account for the factor 25:\[(y^2 - 4y + 4 - 4) = 25(y^2 - 4y + 4 - 4) = 25(y - 2)^2 - 100\].Modify original equation:\[x^2 - 8x + 16 - 16 + 25((y - 2)^2 - 4) + 91 = 0\].

Step 3: Simplify and Arrange Properly

After completing the square in both \(x\) and \(y\) portions, the equation becomes: \[(x - 4)^2 - 16 + 25((y - 2)^2 - 4) + 91 = 0\].Simplify:\[(x - 4)^2 + 25(y - 2)^2 - 100 = -75\].Re-arrange:\[(x - 4)^2 + 25(y - 2)^2 = 100 - 75\].Simplified to:\[(x - 4)^2 + 25(y - 2)^2 = 25\].

Step 4: Write in Standard Form of an Ellipse

Simplify to standard form \(\frac{(x - 4)^2}{25} + \frac{(y - 2)^2}{1} = 1\). Identify \(a\), \(b\), and \(c\): Compare terms with standard ellipse form \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here \(a^2 = 25\), so \(a = 5\), and \(b^2 = 1\), so \(b = 1\).

Step 5: Calculate the Foci Distance

Use the formula for eccentricity \(c^2 = a^2 - b^2\) to find the distance \(c\):\[c^2 = 25 - 1\] \[c^2 = 24\] \[c = \sqrt{24} = 2\sqrt{6}\].

Step 6: Identify the Foci

Since \(a > b\), it's a horizontally oriented ellipse. The foci are along the horizontal axis centered at \(h, k\): \((h \pm c, k) = (4 \pm 2\sqrt{6}, 2)\).Therefore, the foci are at \((4 + 2\sqrt{6}, 2)\) and \((4 - 2\sqrt{6}, 2)\).

Key Concepts

Completing the Square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This is particularly helpful when dealing with conic sections, such as ellipses, to simplify and identify their standard forms.
Given the equation from the exercise, we have terms in both \(x\) and \(y\). The goal is to rearrange and complete the square for each variable separately.
The original equation is \(x^2 - 8x + 25y^2 - 100y + 91 = 0\).

• First, focus on the \(x\) terms: \(x^2 - 8x\).
• Take half of the coefficient of \(x\), which is \(-8\). Half of \(-8\) is \(-4\).
• Square \(-4\) to get \(16\).
• Add and subtract \(16\) around the \(x\) terms: \(x^2 - 8x + 16 - 16\).
• For the \(y\) terms, \(25y^2 - 100y\), first factor out \(25\): \(25(y^2 - 4y)\).
• Complete the square inside the bracket: Half of \(-4\) is \(-2\), and \(-2\) squared is \(4\).
• Add and subtract \(4\) inside the parenthesis: \(25(y^2 - 4y + 4 - 4)\).

This results in a transformed equation that will help us identify the characteristics of the ellipse more easily.

Standard Form of Ellipse

The standard form of an ellipse's equation helps to clearly define its size and shape, as well as its orientation on the coordinate plane. It's expressed as: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here's how to get it from a completed square form:
After completing the square for the original ellipse equation, we simplified it to:\[(x - 4)^2 + 25(y - 2)^2 = 25\]The next step is to express the equation to match the standard form:

• Divide each term by the constant \(25\) to normalize the equation: \[\frac{(x - 4)^2}{25} + \frac{(y - 2)^2}{1} = 1\]
• Here, the center of the ellipse \((h, k)\) is \((4, 2)\), and the lengths \(a\) and \(b\) are derived from the denominators: \(a^2 = 25\), so \(a = 5\); \(b^2 = 1\), so \(b = 1\).

Appreciating this form clarifies not only the size but also the orientation of the ellipse.

Eccentricity of Ellipse

Eccentricity is a measure of how much an ellipse deviates from being a circle. A circle's eccentricity is 0, while an ellipse's is between 0 and 1. The formula to find the eccentricity \(e\) involves \(a\), \(b\), and \(c\), where\(c\) is the focal distance:\[ e = \frac{c}{a} \]Here's how to find \(c\) in our scenario:
Using the formula:\[c^2 = a^2 - b^2\], we can determine \(c\) for the ellipse:

• Given \(a^2 = 25\) and \(b^2 = 1\), we find \(c^2 = 25 - 1 = 24\).
• Taking the square root gives \(c = \sqrt{24} = 2\sqrt{6}\).

The eccentricity is calculated as \(e = \frac{2\sqrt{6}}{5}\). This tells us how elongated the ellipse is compared to a circle.

Foci of Ellipse

The foci (plural of focus) are two distinct points on the major axis of an ellipse. The distance to each focus from the center is denoted as \(c\), and they are crucial in defining the shape of the ellipse.
Finding the foci involves the parameters of the standard equation:

• Since \(a > b\), we know the ellipse is horizontal.
• The foci will be located horizontally from the center \((h, k)\), which here is \((4, 2)\).
• Using \(c = 2\sqrt{6}\), we find the foci at \((h \, \pm c, k)\).
• Thus, the coordinates of the foci are \((4 + 2\sqrt{6}, 2)\) and \((4 - 2\sqrt{6}, 2)\).

Remember, the further apart the foci, the more elongated the ellipse will appear.