Problem 45 - Precalculus: Mathematics for Calculus

Question

Finding the Equation of an Ellipse Find an equation for the ellipse that satisfies the given conditions. Length of major axis: \(4,\) length of minor axis: \(2,\) foci on \(y\) -axis

Step-by-Step Solution

Answer

The equation of the ellipse is \(\frac{x^2}{1} + \frac{y^2}{4} = 1\).

Step 1: Identify the Orientation

Since the foci are on the y-axis, this indicates a vertical ellipse. The major axis is vertical, and the equation for an ellipse centered at the origin with a vertical major axis is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \), where \(a > b\).

Step 2: Determine the Major and Minor Axes Values

Given the length of the major axis is 4, we have \(2a = 4\) which simplifies to \(a = 2\). Given the length of the minor axis is 2, we have \(2b = 2\) which simplifies to \(b = 1\).

Step 3: Calculate the Distance to the Foci

For an ellipse, the distance from the center to each focus is \(c\) where \(c^2 = a^2 - b^2\). Substitute \(a = 2\) and \(b = 1\) into the formula: \(c^2 = 2^2 - 1^2 = 4 - 1 = 3\) so \(c = \sqrt{3}\).

Step 4: Write the Ellipse Equation

Substitute \(a = 2\) and \(b = 1\) into the formula \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\). So, the equation becomes \(\frac{x^2}{1^2} + \frac{y^2}{2^2} = 1\), which simplifies to \(\frac{x^2}{1} + \frac{y^2}{4} = 1\).

Key Concepts

Length of Major Axis

In the world of ellipses, the major axis is always the longest diameter, stretching from one end of the ellipse right through the center to the other end. The length of the major axis is remarkably simple to understand as it consists of two times the longer radius of the ellipse, known as the semi-major axis.

• If the major axis is 4 units long, then this length comprises two equal segments on either side of the center.
• Each segment is known as the semi-major axis, so we express it as "2a". Hence, with a length of 4, the semi-major axis, or "a," is 2 units.

With the major axis along the vertical or horizontal, the ellipse's equation adopts a specific format. If the major axis is vertical, as with our original ellipse, the equation becomes: \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \]. This formula dictates the relationship between the semi-major and semi-minor axes.

Length of Minor Axis

The minor axis of an ellipse represents the shortest diameter, crossing at a right angle to the major axis through the center. This axis acts perpendicular to the longer major axis. Like the major axis, the length of the minor axis contains the shorter radius, known as the semi-minor axis.

• For a minor axis length of 2, similar to our example, the semi-minor axis "b" is half that length, or 1 unit.
• This dimension is crucial for the shape of the ellipse, ensuring its slightly squashed look compared to a circle, which is simply an ellipse with equal major and minor axes.

In the elliptical equation \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \], the "b" value represents the semi-minor axis. This equation outlines how the semi-minor axis influences the ellipse's breadth in its narrowest dimension.

Foci

The concept of foci is unique to ellipses and crucial to understanding their geometric properties. Foci are two fixed points on the interior of an ellipse.

• The combined distance from these points to any point on the ellipse's perimeter remains constant.
• For an ellipse centered at the origin, the foci are located along the major axis.

The distance from the center to each focus is determined by the letter "c". It's calculated using the equation: \[ c^2 = a^2 - b^2 \]In our specific example, "a" is 2, and "b" is 1. Thus, substituting these values results in: \[ c^2 = 2^2 - 1^2 = 3 \] This gives us \(c = \sqrt{3} \), indicating that the foci are \( \sqrt{3} \) units away from the center along the y-axis. This property ensures the ellipse remains in its beautifully stretched form.