Problem 37 - Precalculus Mathematics for Calculus

Question

Find an equation for the ellipse that satisfies the given conditions. Foci: \((0, \pm 2),\) length of minor axis: 6

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{9} + \frac{y^2}{13} = 1\).

Step 1: Understand the Problem

We need to find the equation of an ellipse with given foci and a specified length for the minor axis.

Step 2: Identify the Ellipse Configuration

The foci are \( (0, \, \pm 2) \), indicating a vertical orientation because the y-coordinates change.

Step 3: Determine the Distance Between Foci

The distance between the foci is 4, as they are at \((0, \, 2)\) and \((0, \, -2)\). The formula for the focal distance is \(2c = \) distance between foci. So, \(2c = 4 \), hence \(c = 2\).

Step 4: Analyze Minor Axis Length

The minor axis length is given as 6. The minor axis length is \(2b\), with \(b = 3\).

Step 5: Relation Between a, b, and c for Ellipses

In a vertical ellipse, the relationship is \(c^2 = a^2 - b^2\). We already know \(c = 2\) and \(b = 3\), so we can solve for \(a\).

Step 6: Solve for a

Substitute the values: \(2^2 = a^2 - 3^2\) This simplifies to \(4 = a^2 - 9\). Add 9 to both sides to get: \(a^2 = 13\).

Step 7: Write the Equation for the Ellipse

Since the ellipse is vertical, the standard form of the equation is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\). Substituting, \(\frac{x^2}{3^2} + \frac{y^2}{13} = 1\). Therefore, the equation becomes \[ \frac{x^2}{9} + \frac{y^2}{13} = 1 \].

Key Concepts

Foci of an Ellipse

The foci of an ellipse are two fixed points located inside the ellipse. These points help determine the shape and orientation of the ellipse. For any point on the ellipse, the sum of the distances from that point to each focus is constant. The position of the foci (

• For horizontal ellipses, the foci are positioned along the x-axis, meaning their coordinates are \(( ext{h} \pm c, ext{k})\).
• For vertical ellipses, the foci are along the y-axis, like in this case, where the foci are \((0, \pm 2)\).

Here, the foci tell us that the ellipse has a vertical orientation. The distance between these foci is critical, as it helps us calculate the eccentricity of the ellipse, which describes its "stretch." Understanding the position and calculation of the foci is essential in writing the equation of an ellipse.

Minor Axis Length

The minor axis of an ellipse refers to the shortest diameter, which runs perpendicular to the major axis. It cuts through the center, or midpoint, of the ellipse. The length of the minor axis is a crucial feature in determining the size and aspect of the ellipse. The minor axis length is given as 6 in this problem. Since the minor axis relates to the value \(2b\) (representing the total length), we can calculate \(b\) as:- \(b = \frac{6}{2} = 3\). This value of \(b\) will be used in the standard form of the ellipse's equation.By knowing the length of the minor axis, we gain insight into the proportions of the ellipse, allowing it to have a clearer and mathematically accurate visual representation.

Ellipse Standard Form

When it comes to finding the equation of an ellipse, you'll usually work with the ellipse standard form. This formula helps describe the ellipse algebraically and varies depending on the ellipse's orientation. For a vertical ellipse, like the one in this example, the standard form is:\[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]In this case:

• \(b^2\) is derived from the minor axis, giving us \(b^2 = 9\).
• To find \(a^2\), we use the relationship \(c^2 = a^2 - b^2\). With \(c = 2\) and \(b = 3\), you can solve for \(a^2\):
• \(4 = a^2 - 9\), so adding 9 to both sides results in \(a^2 = 13\).

Putting it all together, the equation of the ellipse becomes:\[ \frac{x^2}{9} + \frac{y^2}{13} = 1 \].This form not only captures the essence of the ellipse but also explains its dimensions and orientation in the coordinate plane, making it an essential tool for understanding and working with ellipses.