Problem 61 - Algebra 2

Question

Write an equation of an ellipse in standard form with center at the origin and with the given characteristics. focus \((3,0), x\) -intercept \(-6\)

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{36} + \frac{y^2}{27} = 1 \)

Step 1: Identify and calculate the semi-major axis

The distance from the center of the ellipse to either of its x-intercepts is the semi-major axis length, a. We are given that one x-intercept is -6, so \( a = 6 \)

Step 2: Identify and calculate the distance from the center to the focus

The distance from the center of the ellipse to either of its foci along the major axis is given as 3. Denote this length as c. Therefore, \( c = 3 \)

Step 3: Use the relationship between a, b, and c to find the semi-minor axis

The semi-minor axis length, b, can be found using the relationship in ellipses that \( a^2 = b^2 + c^2 \). Substituting the known values, we have \( b = \sqrt{a^2 - c^2} = \sqrt{6^2 - 3^2} = \sqrt{27} \)

Step 4: Write the equation of the ellipse

The standard form equation of the ellipse with a horizontal major axis is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substituting for a and b, we get \( \frac{x^2}{36} + \frac{y^2}{27} = 1 \)

Key Concepts

Semi-Major Axis

The semi-major axis is a crucial element of an ellipse. It acts as the longest radius of the ellipse, stretching from the center to the furthest edge along its longer axis. In simple terms, it's half the distance across the widest part of the ellipse.
In our problem, the semi-major axis is identified using the ellipse's x-intercept. Since the ellipse is centered at the origin and has an x-intercept at -6, we know the entire x-intercept spans from -6 to 6. Therefore, the semi-major axis length is the absolute value of -6, which is 6. This distance determines the ellipse's width along the horizontal direction.
Understanding the semi-major axis helps us see how stretched the ellipse is, and it's always related to the length of the minor axis and the focus points.

Semi-Minor Axis

The semi-minor axis is the shorter radius of an ellipse, perpendicular to the semi-major axis. If you imagine the ellipse as a stretched circle, the semi-minor axis would represent the lesser stretch.
To calculate the semi-minor axis, we utilize an important ellipse property: the Pythagorean relationship between the semi-major axis \(a\), semi-minor axis \(b\), and the distance to the focus \(c\). This relationship states that \(a^2 = b^2 + c^2\).
With \(a = 6\) and \(c = 3\), we can find \(b\) using the formula \(b = \sqrt{a^2 - c^2} = \sqrt{6^2 - 3^2} = \sqrt{27}\). Calculating this gives us \(b \approx 5.2\). The semi-minor axis, therefore, shows us how tightly the ellipse is compressed across its shorter width.

Standard Form Equation

The standard form equation of an ellipse allows us to algebraically define the shape and size of the ellipse on a 2D plane. This equation relates the x and y coordinates of any point on the ellipse to its semi-major and semi-minor axes.
The general standard form for an ellipse with a horizontal major axis looks like this: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). It represents all points (x, y) that constitute the ellipse.
Given our earlier calculations, where \(a = 6\) and \(b = \sqrt{27}\), the equation becomes:

• \(\frac{x^2}{36} + \frac{y^2}{27} = 1\)

This equation tells us everything we need to determine the ellipse's shape and position, with its center at the origin and its major and minor axes properly proportioned.