Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. \(4 x^{2}+40 x+25 y^{2}-100 y+100=0\)

Step-by-Step Solution

Answer

The standard form is \(\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1\). Major axis endpoints: (0,2) and (-10,2), minor axis endpoints: (-5,4) and (-5,0), foci: approximately (-0.42, 2) and (-9.58, 2).

Step 1: Rearrange the Equation

Group the equation by variable terms as follows:\[4x^2 + 40x + 25y^2 - 100y = -100 \]

Step 2: Complete the Square for x-terms

Factor out the 4 from the x-terms:\[4(x^2 + 10x) \]To complete the square:\[4(x^2 + 10x + 25 - 25)\]Rewrite:\[4((x+5)^2 - 25)\]

Step 3: Complete the Square for y-terms

Factor out the 25 from the y-terms:\[25(y^2 - 4y) \]To complete the square:\[25(y^2 - 4y + 4 - 4) \]Rewrite:\[25((y-2)^2 - 4)\]

Step 4: Simplify the Expression

Substitute the completed squares into the grouped equation:\[4((x+5)^2 - 25) + 25((y-2)^2 - 4) = -100\]Distribute and combine like terms:\[4(x+5)^2 - 100 + 25(y-2)^2 - 100 = -100\]\[4(x+5)^2 + 25(y-2)^2 = 100\]

Step 5: Divide to Achieve Standard Form

Divide every term by 100 to get the standard form:\[\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1 \]This is the standard form of the ellipse \(\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1\).

Step 6: Identify the Components of the Ellipse

From the standard form \(\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1\):- Center: \((-5, 2)\)- Semi-major axis \(a = 5\) (since 25 is larger)- Semi-minor axis \(b = 2\)- Foci distance \(c\): \(c = \sqrt{25 - 4} = \sqrt{21} \approx 4.58\)

Step 7: Identify Endpoints of Axes and Foci

- Major axis endpoints: \((-5 \pm 5, 2) = (0, 2)\) and \((-10, 2)\)- Minor axis endpoints: \((-5, 2 \pm 2) = (-5, 4)\) and \((-5, 0)\)- Foci: \((-5 \pm \sqrt{21}, 2)\) which approximately are \((-0.42, 2)\) and \((-9.58, 2)\)

Key Concepts

Standard Form

The standard form of an ellipse's equation is essential to easily identifying its properties. The typical format for the standard form of an ellipse is:\[\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1\]Here,

• \((h, k)\) is the center of the ellipse.
• \(a\) is the semi-major axis length.
• \(b\) is the semi-minor axis length.

This form allows us to swiftly determine the center, axes, and other components of the ellipse. In our exercise, after simplifying and rearranging the given equation, we obtain the standard form:\[\frac{(x+5)^2}{25} + \frac{(y-2)^2}{4} = 1\]This indicates the center of the ellipse is at \((-5, 2)\), the semi-major axis is 5, and the semi-minor axis is 2.

Major Axis Endpoints

The major axis of an ellipse is the longest diameter that passes through the center. For ellipses, the major axis corresponds to the larger denominator in the standard form's equation:

• Here, 25 is larger than 4.
• So, the major axis is parallel to the x-axis.

To find the endpoints of the major axis, we use:- Center coordinates: \((-5, 2)\)- Semi-major axis length, \(a = 5\)Endpoints of the major axis are found by altering the x-coordinate:

• Add and subtract the semi-major axis length to the x-coordinate of the center.

Thus, the major axis endpoints are calculated as:\((-5 + 5, 2) = (0, 2)\) and \((-5 - 5, 2) = (-10, 2)\).

Minor Axis Endpoints

The minor axis is the shortest diameter of the ellipse, perpendicular to the major axis. Using our standard form, the denominator 4, corresponding to the minor axis, is smaller than 25:

• This indicates that the minor axis is aligned with the y-axis.

For the minor axis endpoints, we will:

• Retain the x-coordinate from the ellipse's center.
• Vary the y-coordinate by the semi-minor axis length, \(b = 2\).

Thus, the minor axis endpoints are derived as:\((-5, 2 + 2) = (-5, 4)\) and \((-5, 2 - 2) = (-5, 0)\).

Foci

The foci of an ellipse are two points located along the major axis such that the sum of the distances from the foci to any point on the ellipse remains constant. To determine the foci positions, the distance \(c\) from the center to each focus needs to be calculated. This distance is calculated using the formula:\[c = \sqrt{a^2 - b^2}\]Here:

• The semi-major axis \(a = 5\)
• The semi-minor axis \(b = 2\)

Plug these into the formula:\[c = \sqrt{25 - 4} = \sqrt{21} \approx 4.58\]The foci are situated at \((-5 \pm 4.58, 2)\). Approximating further gives the foci coordinates as:

• \((-0.42, 2)\) and \((-9.58, 2)\)

Understanding the location of foci gives insight into the geometric and reflective properties of the ellipse.