Problem 37 - Algebra and Trigonometry

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: Identify the Standard Form of an Ellipse

An ellipse centered at the origin \(0,0\) with a vertical major axis has the standard form\( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Here, \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

Step 2: Discuss the Foci Relationship

For an ellipse centered at the origin with foci on the y-axis, the distance from the center to a focus is given by \(c\), where \(c^2 = a^2 - b^2\). Here, the foci are at \( (0, \pm 2)\), so \(c = 2\).

Step 3: Determine the Length of the Minor Axis

The total length of the minor axis is given, which is 6. Therefore, the semi-minor axis \(b\) is half of this length: \(b = 3\).

Step 4: Calculate the Semi-Major Axis

Use the foci relationship \(c^2 = a^2 - b^2\) to find \(a\). Substitute \(c = 2\) and \(b = 3\): \(2^2 = a^2 - 3^2\). Simplify to find \(4 = a^2 - 9\), so \(a^2 = 13\). Thus, \(a = \sqrt{13}\).

Step 5: Write the Equation of the Ellipse

Substitute \(a\) and \(b\) into the standard form equation. The equation is \( \frac{x^2}{9} + \frac{y^2}{13} = 1 \), with \(a^2 = 13\) and \(b^2 = 9\).

Key Concepts

Foci

In an ellipse, the foci (plural of focus) are two points located along the major axis. These points are essential in defining the shape and properties of the ellipse. For any point on the ellipse, the sum of the distances to the two foci is constant. This property uniquely defines the ellipse and is crucial for understanding its structure.
In the given exercise, the foci are positioned at

• he (0, +2) and (0, -2) along the y-axis.

The positions suggest the ellipse has a vertical major axis. Determining the location of the foci helps us calculate other parameters, like the semi-major and semi-minor axes. The distance from the center of the ellipse to each focus is denoted as \( c \). In this case, \( c = 2 \). This value is later used alongside the semi-minor axis to find the length of the semi-major axis, but more on that in the following sections.

Semi-Major Axis

The semi-major axis is the longest radius of an ellipse and stretches from the center to the furthest point on the curve. It lies along the ellipse's major axis—a distinguishing feature of an ellipse helping to differentiate it from a circle.
The length of the semi-major axis is symbolized by \( a \). To find this length, we utilize the relationship involving the foci and the semi-minor axis in the formula \( c^2 = a^2 - b^2 \). Given in our exercise:

• the foci distance \( c = 2 \),
• the semi-minor axis \( b = 3 \),

we can solve \( 2^2 = a^2 - 3^2 \), which simplifies to \( a^2 = 13 \). Thus, the length of the semi-major axis \( a \) is \( \sqrt{13} \). Knowing \( a \), we can formulate the ellipse equation and understand its spatial proportions.

Semi-Minor Axis

The semi-minor axis is the shorter radius of an ellipse, stretching from the center to the shortest point on its curve. It is located along the axis that is perpendicular to the major axis. Represented by \( b \), this measure helps define the curve's width and is equally important in forming the complete ellipse equation.
In the problem provided, the full minor axis is noted to measure 6. Therefore, dividing this total length by 2 gives us the semi-minor axis:

• \( b = 3 \).

We use the value of \( b \) when applying the standard form equation of the ellipse \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Incorporating \( b = 3 \), we substitute into \( b^2 = 9 \). This detail ties with the foci and semi-major axis to flesh out the ellipse's final equation, showing its characteristics and bounds on a coordinate plane.