Problem 41 - Precalculus Mathematics for Calculus

Question

Find an equation for the ellipse that satisfies the given conditions. Length of major axis: \(10 .\) foci on \(x\) -axis, ellipse passes through the point \((\sqrt{5}, 2)\)

Step-by-Step Solution

Answer

Equation: \( \frac{x^2}{25} + \frac{y^2}{5} = 1 \).

Step 1: Understanding the Problem

To find the equation of an ellipse with given conditions, start by identifying key information: the length of the major axis, the orientation (foci on the x-axis), and a specific point the ellipse passes through.

Step 2: Identify Ellipse Parameters

Since the foci are on the x-axis, the standard form of the ellipse equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). The center \((h,k)\) is at the origin \((0, 0)\) because no specific center was given.

Step 3: Determine Semi-Major Axis

The length of the major axis is 10, which implies the semi-major axis \(a\) is half of this, so \(a = 5\).

Step 4: Set up the Equation

Use the standard form of the ellipse equation with the center at the origin: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substituting \(a = 5\), we get \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \).

Step 5: Use the Given Point to Find \(b^2\)

The point \((\sqrt{5}, 2)\) lies on the ellipse, so substitute \(x = \sqrt{5}\) and \(y = 2\) into the equation: \( \frac{(\sqrt{5})^2}{25} + \frac{2^2}{b^2} = 1 \). This simplifies to \( \frac{5}{25} + \frac{4}{b^2} = 1 \).

Step 6: Solve for \(b^2\)

Solve the equation \( \frac{5}{25} + \frac{4}{b^2} = 1 \) to find \(b^2\). Subtract \(\frac{5}{25} \) from both sides to get \( \frac{4}{b^2} = \frac{20}{25} = \frac{4}{5} \). Thus, \(b^2 = 5\).

Step 7: Write the Complete Ellipse Equation

With \(a^2 = 25\) and \(b^2 = 5\), the complete equation is \( \frac{x^2}{25} + \frac{y^2}{5} = 1 \).

Key Concepts

Semi-Major Axis

In an ellipse, one of the fundamental components is the semi-major axis. It represents half of the longest diameter of the ellipse, known as the major axis. In our problem, the length of the major axis is given as 10.
This means that the semi-major axis, denoted as \(a\), is half of this value. So in this case, \(a = \frac{10}{2} = 5\).
The semi-major axis determines the extent of the ellipse along its broader dimension.Understanding the concept of the semi-major axis is key because it helps in formulating the ellipse's equation and assessing its shape. The greater the semi-major axis, the more elongated the ellipse will be along that direction.

Foci

An ellipse has two focal points, known as foci. These points lie along the major axis, equidistant from the center of the ellipse. The presence of the foci is pivotal to the definition and properties of an ellipse: the sum of the distances from any point on the ellipse to the two foci is constant.In this exercise, since the ellipse's foci are located on the \(x\)-axis, it implies a certain perception and tilting of the ellipse. The foci are closer to the center in more circular ellipses and further apart in more elongated ones.Calculating the exact position of these foci involves understanding the relationship \(c^2 = a^2 - b^2\) where \(c\) is the distance from the center to each focus, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis.

Standard Form of Ellipse

The standard form of an ellipse's equation offers a structured way to represent its shape and position. When centered at the origin, this equation is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).Here, \((h, k)\) represents the center of the ellipse. In our exercise, the center is at the origin \((0, 0)\), so the equation simplifies to \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Substituting the calculated \(a^2 = 25\) and \(b^2 = 5\), the ellipse's equation becomes \( \frac{x^2}{25} + \frac{y^2}{5} = 1 \).This equation is vital as it helps visualize how the ellipse is oriented and scaled within the xy-plane, especially when it passes through specific points, such as \((\sqrt{5}, 2)\).