Problem 37 - Algebra and Trigonometry

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

The equation of the ellipse is \( \frac{x^2}{25} + \frac{y^2}{5} = 1 \).

Step 1: Identify the general equation

An ellipse aligned with the coordinate axes has the general equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Since the foci are on the \(x\)-axis, we conclude that the ellipse's major axis is horizontal.

Step 2: Determine semi-major axis

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

Step 3: Use the point to find the semi-minor axis

Plug the point (\(\sqrt{5}, 2\)) into the ellipse equation \( \frac{x^2}{25} + \frac{y^2}{b^2} = 1 \).

Step 4: Substitute and solve for \(b^2\)

Substitute \(x = \sqrt{5}\) and \(y = 2\) into the equation: \( \frac{(\sqrt{5})^2}{25} + \frac{2^2}{b^2} = 1 \). Simplifying gives \( \frac{5}{25} + \frac{4}{b^2} = 1 \).

Step 5: Simplify the equation

Solve \( \frac{1}{5} + \frac{4}{b^2} = 1 \). This can be rewritten as \( \frac{4}{b^2} = \frac{4}{5} \).

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

Rearrange to find \( b^2 = 5 \).

Step 7: Write the final equation

Substitute \(a^2 = 25\) and \(b^2 = 5\) back into the ellipse equation: \( \frac{x^2}{25} + \frac{y^2}{5} = 1 \).

Key Concepts

Semi-Major Axis

The semi-major axis of an ellipse is one of its most important features. It represents half the length of the major axis, which is the longest diameter of the ellipse. In an ellipse, the major axis runs through both foci, thus determining the stretch of the ellipse along its longest direction. For the ellipse discussed in our original exercise, the major axis is 10 units long. To find the semi-major axis, we divide this length by two. As a consequence, our semi-major axis is 5 units long. This means that horizontally, from the center of the ellipse to its farthest edge along the x-axis, we measure 5 units. Thus, the value of the semi-major axis directly impacts the shape and size of the ellipse.

Semi-Minor Axis

The semi-minor axis is another vital component of an ellipse. While the semi-major axis is the longest radius, the semi-minor axis is the shortest. It measures half the length of the shortest axis, which is perpendicular to the major axis.In our problem, after using the point \(\sqrt{5}, 2\) and the formula for an ellipse, we find that the semi-minor axis has a length that satisfies the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Upon inserting the point into the equation, we solved for \(b^2\), resulting in \(b = \sqrt{5}\). This indicates that the semi-minor axis is roughly 2.23 units long, revealing how much shorter it is compared to the semi-major axis and contributing to the oval shape of the ellipse.

Foci of Ellipse

The foci of an ellipse are two fixed points located along the major axis. These points hold a unique property: any point on the ellipse has a sum of distances to the two foci that is constant.In our given problem, the foci of the ellipse are positioned on the x-axis, which aligns with our determination that the major axis is horizontal. The distance from the center of the ellipse to each focus is calculated using the equation \( c = \sqrt{a^2 - b^2} \).Using the values of \(a^2 = 25\) and \(b^2 = 5\), we find \( c = \sqrt{25 - 5} = \sqrt{20}\), which simplifies to approximately 4.47 units. These calculations enable us to place the foci symmetrically along the x-axis, providing insights into the elliptical properties of distance and shape.