Problem 38 - A Graphical Approach to Precalculus with Limits

Question

Find an equation for each ellipse. Foci at \((-3,-3)\) and \((7,-3) ;\) the point \((2,1)\) on ellipse

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{(x-2)^2}{41} + \frac{(y+3)^2}{16} = 1 \).

Step 1: Determine the Center

The foci of the ellipse are given as \((-3, -3)\) and \((7, -3)\). The center of the ellipse is the midpoint of the segment joining the foci. To find the midpoint, use the formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\).Calculate the midpoint:\[\left( \frac{-3 + 7}{2}, \frac{-3 + (-3)}{2} \right) = \left( \frac{4}{2}, \frac{-6}{2} \right) = (2, -3)\]Thus, the center of the ellipse is \((2, -3)\).

Step 2: Establish the Orientation of the Ellipse

Since the foci are aligned horizontally, the major axis is horizontal.This implies that the equation of the ellipse will be in the form: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where \((h, k)\) is the center and \(a > b\).

Step 3: Calculate Distance Between Foci and Find \(c\)

The distance between the foci is the distance between the points \((-3, -3)\) and \((7, -3)\).Calculate the distance (denoted as \(2c\)):\[ 2c = |7 - (-3)| = 7 + 3 = 10 \]Thus, \(c = \frac{10}{2} = 5\).

Step 4: Use the Given Point to Find \(b^2\)

The given point \((2, 1)\) lies on the ellipse. Substitute this point and center into the ellipse equation:\[ \frac{(2-2)^2}{a^2} + \frac{(1 - (-3))^2}{b^2} = 1 \]This simplifies to:\[ \frac{0}{a^2} + \frac{16}{b^2} = 1 \]Thus, \(\frac{16}{b^2} = 1\), leading to \(b^2 = 16\).

Step 5: Use the Relationship \(a^2 = b^2 + c^2\) to Find \(a^2\)

Using the formula \(a^2 = b^2 + c^2\), calculate \(a^2\):Plug in \(b^2 = 16\) and \(c = 5\) (so \(c^2 = 25\)) to find:\[a^2 = 16 + 25 = 41\]

Step 6: Write the Equation of the Ellipse

Substitute the calculated values into the ellipse equation:The center \((h, k)\) is \((2, -3)\), \(a^2 = 41\), and \(b^2 = 16\).Thus, the equation is:\[ \frac{(x-2)^2}{41} + \frac{(y+3)^2}{16} = 1 \]

Key Concepts

Foci

In the context of ellipses, the foci are two special points. They impact the shape and positioning of the ellipse. Imagine an ellipse as a stretched circle with two internal focal points. These points, known as foci (singular: focus), determine the ellipse's precise orientation and size.

In an ellipse, the sum of the distances from any point on the ellipse to each focus is constant. This property allows us to define an ellipse geometrically. Given the foci at \((-3, -3)\) and \((7, -3)\), you can determine the major axis as well as the orientation of the ellipse.

• The foci are horizontally aligned, indicating that the major axis is along the x-axis.
• The formula mentioned in our example for the orientation is affected by these foci locations.

Understanding the role of foci helps in deriving the equation of the ellipse and predicting its behavior in a plane.

Midpoint Formula

The midpoint formula is crucial for finding the center of an ellipse when the coordinates of the foci are known. This formula is a simple yet powerful tool used in geometry.

Let's break down the midpoint formula: It calculates the point halfway between two other points, helping to determine the centroid of an object. For an ellipse, it's essential for determining the center from the foci coordinates. If you have two points \((x_1, y_1)\) and \((x_2, y_2)\), the formula calculates:

• Midpoint = \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

In our problem, we used this formula with the foci \((-3, -3)\) and \((7, -3)\), giving us the ellipse center at \((2, -3)\). Knowing how to use the midpoint formula is essential when dealing with conic sections like ellipses.

Major Axis

The major axis of an ellipse is the longest diameter that passes through both foci and the center of the ellipse. It sets the direction and length of the ellipse's largest expansion.

In our exercise, the major axis is horizontal because the foci's y-coordinates are identical. This means the ellipse is wider horizontally than vertically. Understanding the major axis is vital for establishing the equation of an ellipse.

• Major axis always passes through the center.
• The length of the major axis corresponds to the variable \(a\) when forming the ellipse equation.

To find \(a\), we use the relation \(a^2 = b^2 + c^2\), where \(c\) is the distance from the center to a focus, and \(b\) relates to the minor axis. Knowing the length and orientation of the major axis guides us in defining the ellipse's equation.

Ellipse Properties

Ellipses have fascinating properties that make them unique among other geometric figures. These properties define their shape, size, and equation.

When discussing properties, the equation of an ellipse holds critical importance. In our exercise, the equation was:\[ \frac{(x-2)^2}{41} + \frac{(y+3)^2}{16} = 1 \]Each part of this equation contributes to the overall properties of the ellipse.

• The fraction formulas show that the ellipse's center is at \((2, -3)\).
• The denominators \(41\) and \(16\) relate to the semi-major axis and semi-minor axis. \(a^2\) and \(b^2\) are these values, respectively.
• Moreover, due to the horizontal major axis, the ellipse spreads more along the x-axis.

Understanding these unique properties gives depth to how ellipses behave and allows you to analyze their structures in varied contexts. Ellipses are not just simple elongated circles; they possess symmetry and precision.