Problem 15 - 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. \(\frac{(x-2)^{2}}{49}+\frac{(y-4)^{2}}{25}=1\)

Step-by-Step Solution

Answer

The ellipse's equation is in standard form: \( \frac{(x-2)^2}{49} + \frac{(y-4)^2}{25} = 1 \). Major axis endpoints: (9,4) and (-5,4). Minor axis endpoints: (2,9) and (2,-1). Foci: (6.9,4) and (-2.9,4).

Step 1: Identify the Standard Form

The equation given is \( \frac{(x-2)^{2}}{49} + \frac{(y-4)^{2}}{25} = 1 \), which is already in the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \((h, k) = (2, 4)\), \(a^2 = 49\), and \(b^2 = 25\). Because \(a^2 > b^2\), the major axis is horizontal.

Step 2: Determine the Values of a and b

Calculate \(a\) and \(b\) by taking the square roots of \(a^2\) and \(b^2\). We get \(a = \sqrt{49} = 7\) and \(b = \sqrt{25} = 5\).

Step 3: Identify the Center of the Ellipse

The center \((h, k)\) is given by \((2, 4)\). This is the midpoint of both the major and minor axes.

Step 4: Find the End Points of the Major Axis

The major axis is horizontal, so apply this to the center \((2, 4)\). The endpoints are \((2 \pm a, 4)\), leading to \((2+7, 4) = (9, 4)\) and \((2-7, 4) = (-5, 4)\).

Step 5: Find the End Points of the Minor Axis

The minor axis is vertical, so apply this to the center \((2, 4)\). The endpoints are \((2, 4 \pm b)\), leading to \((2, 4+5) = (2, 9)\) and \((2, 4-5) = (2, -1)\).

Step 6: Calculate the Foci

The distance of the foci from the center is given by \(c = \sqrt{a^2 - b^2} = \sqrt{49 - 25} = \sqrt{24}\). Therefore, the foci are located at \((2 \pm \sqrt{24}, 4)\), simplifying to \((2+\sqrt{24}, 4)\) and \((2-\sqrt{24}, 4)\), which approximately gives \((6.9, 4)\) and \((-2.9, 4)\).

Key Concepts

Ellipse Standard Form

The standard form of an ellipse equation is a crucial starting point in understanding ellipses. It's presented as \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] where

• \((h, k)\) designates the center of the ellipse
• \(a^2\) and \(b^2\) represent the squared lengths of the semi-major and semi-minor axes, respectively.

The given equation \[\frac{(x-2)^2}{49} + \frac{(y-4)^2}{25} = 1\] indicates that the center is at \((h, k) = (2, 4)\). This format simplifies identifying whether the major axis is horizontal or vertical.
By comparing values \(a^2 = 49\) and \(b^2 = 25\), we learn that \(a^2 > b^2\), confirming a horizontal major axis. Recognizing the correct standard form elements allows us to systematically find critical features like axes endpoints and foci.

Major and Minor Axes

In an ellipse's geometry, identifying the major and minor axes is key to understanding its shape and orientation. For the equation \[\frac{(x-2)^2}{49} + \frac{(y-4)^2}{25} = 1\], the values \(a\) and \(b\) are found by

• Calculating \(a = \sqrt{49} = 7\), so the semi-major axis length is 7 units.
• Calculating \(b = \sqrt{25} = 5\), so the semi-minor axis length is 5 units.

The center at \((2, 4)\) helps position these axes. The endpoints of the major axis, being horizontal, are at

• \((2+a, 4) = (9, 4)\)
• \((2-a, 4) = (-5, 4)\)

Meanwhile, the vertical minor axis ends at

• \((2, 4+b) = (2, 9)\)
• \((2, 4-b) = (2, -1)\)

Understanding these axes not only constructs the skeleton of an ellipse but also gives insight into the ellipse's extent and direction.

Finding Foci

The foci of an ellipse hold a special significance in its geometric properties and require careful calculation. Here, they determine the ellipse's shape by ensuring the constant sum of the distances to any point on its outline.
To find the foci for the equation \[\frac{(x-2)^2}{49} + \frac{(y-4)^2}{25} = 1\], use the formula \(c = \sqrt{a^2 - b^2}\), where

• \(a^2 = 49\)
• \(b^2 = 25\)

Thus, \[c = \sqrt{49 - 25} = \sqrt{24}\] approximating to \(c \approx 4.9\).
From the center \((2, 4)\), extrapolate the foci along the horizontal major axis:

• \((2 + \sqrt{24}, 4) \approx (6.9, 4)\)
• \((2 - \sqrt{24}, 4) \approx (-2.9, 4)\)

By understanding the foci's role, you grasp a deeper level of elliptical geometry extending beyond mere visuals.

Geometry of Conic Sections

Ellipses are an integral part of conic sections, a set of curves formed by intersecting a plane with a cone. This intersection creates various shapes, including circles, ellipses, parabolas, and hyperbolas, depending on the angle.
An ellipse, such as the one given by \[\frac{(x-2)^2}{49} + \frac{(y-4)^2}{25} = 1\], is characterized by its irregular curve reflecting symmetry around its axes. Here's how it fits:

• Major and minor axes define its longest and shortest dimensions.
• The foci further elaborate its form by maintaining equidistant sum across its path.

Elliptical geometry is not just mathematical. It appears in planetary orbits, reflecting elliptical paths as planets circle stars. This context reveals how studying ellipses through conic sections offers insight into both abstract math and practical science applications.