Problem 35 - College Algebra

Question

For the following exercises, graph the given ellipses, noting center, vertices, and foci. $$ 81 x^{2}+49 y^{2}=1 $$

Step-by-Step Solution

Answer

Center: (0,0); Vertices: (0, ±1/7); Foci: (0, ±√32/63).

Step 1: Review the Standard Form of an Ellipse

The standard form of an ellipse 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, and \(b\) is the semi-minor axis. An ellipse is horizontal if \(a^2 > b^2\) and vertical if \(b^2 > a^2\).

Step 2: Rewrite the Given Equation in Standard Form

The given equation is \( 81x^2 + 49y^2 = 1 \). To rewrite it in standard form, divide every term by 1 (the right side remains the same). This already achieves the form: \( \frac{x^2}{\frac{1}{81}} + \frac{y^2}{\frac{1}{49}} = 1 \).

Step 3: Identify the Parameters

From the rewritten equation, \(a^2 = \frac{1}{81}\) and \(b^2 = \frac{1}{49}\), so \(a = \frac{1}{9}\) and \(b = \frac{1}{7}\). Since \(a < b\), the ellipse has a vertical orientation. The center is at \((0, 0)\) because there are no \(h\) and \(k\) terms present.

Step 4: Determine the Vertices

For a vertically oriented ellipse, the vertices are located at \((0, \pm b)\). Thus, using \(b = \frac{1}{7}\), the vertices are at \((0, \pm \frac{1}{7})\).

Step 5: Calculate the Foci

For the foci of an ellipse, we use the formula \(c^2 = b^2 - a^2\). Substituting the values gives \(c^2 = \frac{1}{49} - \frac{1}{81}\). Solve for \(c^2\): \(c^2 = \frac{81 - 49}{3969} = \frac{32}{3969}\). Therefore, \(c = \frac{\sqrt{32}}{63}\). The foci are located at \((0, \pm \frac{\sqrt{32}}{63})\).

Step 6: Graph the Ellipse

Plot the center at the origin \((0, 0)\). Mark the vertices at \((0, \frac{1}{7})\) and \((0, -\frac{1}{7})\). The semi-minor axis extends horizontally at \((\pm \frac{1}{9}, 0)\). Plot the foci at \((0, \frac{\sqrt{32}}{63})\) and \((0, -\frac{\sqrt{32}}{63})\). Draw the ellipse to connect these points, ensuring it stretches vertically.

Key Concepts

Standard Form of an Ellipse

When you're dealing with an ellipse, one of the first steps is to express the equation in its standard form. This form makes it easier to identify important aspects of the ellipse, such as its orientation and dimensions. The standard form of an ellipse is:

• \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)

Here, \((h, k)\) represents the center of the ellipse. The values \(a\) and \(b\) denote the lengths of the semi-major and semi-minor axes, respectively. An ellipse is considered horizontal if \(a^2 > b^2\), meaning it stretches more along the x-axis. Conversely, it is vertical if \(b^2 > a^2\), indicating a greater stretch along the y-axis.
Some problems may not be initially presented in this standard form. You might need to manipulate the equation, often by factoring or dividing, to achieve it.
In our exercise, the equation \(81x^2 + 49y^2 = 1\) was transformed into the standard form through simplification, whereby each term was divided by 1. This step then allowed us to properly analyze the ellipse by identifying when \(a\) and \(b\) values suggest a vertical ellipse.

Vertices of an Ellipse

Vertices are key points on an ellipse that define its extent along its major axis. They help us understand the size and orientation of the ellipse. For a vertical ellipse like the one in our exercise, the vertices are found along the y-axis. The general formula to find the vertices is:

• For a vertical orientation: \((h, k \pm b)\)
• For a horizontal orientation: \((h \pm a, k)\)

Where \(h\) and \(k\) are the coordinates of the ellipse's center, and \(a\) and \(b\) are half the lengths of the major and minor axes respectively. The plus-minus sign indicates that the vertices are equidistant above and below (or left and right) of the center.
In the exercise, since the ellipse is vertical and centered at the origin \((0, 0)\), the vertices are at \((0, \pm \frac{1}{7})\). This means the ellipse stretches from just above the origin to just below it, signifying a vertical orientation.

Foci of an Ellipse

The foci (singular: focus) of an ellipse are two fixed points located inside the ellipse. These points are crucial in defining the ellipse's shape, based on the property that the sum of the distances from any point on the ellipse to the foci is constant. To find the foci, you utilize the relationship:

• \(c^2 = b^2 - a^2\) for a vertical ellipse
• \(c^2 = a^2 - b^2\) for a horizontal ellipse

Where \(c\) is the distance from the center of the ellipse to each focus. This distance helps us know where the foci are located compared to the center.
In our exercise, for the vertically-oriented ellipse, we calculated \(c\) using the values \(b^2 = \frac{1}{49}\) and \(a^2 = \frac{1}{81}\). This calculation yielded \(c = \frac{\sqrt{32}}{63}\), so the foci are at \((0, \pm \frac{\sqrt{32}}{63})\). These positions are slightly closer to the origin than the vertices, tucked inside the ellipse along the y-axis.