Problem 26 - Thomas Calculus

Question

Exercises 25 and 26 give information about the foci and vertices of ellipses centered at the origin of the \(x y\) -plane. In each case, find the ellipse's standard-form equation from the given information. $$ \begin{array}{l}{\text { Foci } :(0, \pm 4)} \\ {\text { Vertices: }(0, \pm 5)}\end{array} $$

Step-by-Step Solution

Answer

The standard form equation is \(\frac{x^2}{9} + \frac{y^2}{25} = 1\).

Step 1: Identify Key Parameters

The foci are at \((0, \pm 4)\) and the vertices are at \((0, \pm 5)\). This indicates that the ellipse is vertical, meaning its major axis aligns along the y-axis. For a vertical ellipse centered at the origin, the standard form is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\).

Step 2: Define Parameters and Relationships

For an ellipse, the relationship between the semi-major axis \(a\), semi-minor axis \(b\), and the distance to the foci \(c\) is \(c^2 = a^2 - b^2\). Here, the distance \(c\) of the foci from the center is 4, and the distance \(a\) of the vertices is 5, so \(a = 5\).

Step 3: Calculate Semi-Major Axis Length

Substitute the known values in the foci-vertex relationship: \(c = 4\) and \(a = 5\). We can find the unknown \(b\) using \(c^2 = a^2 - b^2\).

Step 4: Compute and Solve Equation

Substituting the known parameters into the equation: \(4^2 = 5^2 - b^2\), which simplifies to \(16 = 25 - b^2\). Solve this equation for \(b^2\).

Step 5: Conclude with Standard Form Equation

Solving \(16 = 25 - b^2\) gives \(b^2 = 9\). Thus, the standard-form equation of the ellipse \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) is \(\frac{x^2}{9} + \frac{y^2}{25} = 1\).

Key Concepts

Foci of an Ellipse

The foci of an ellipse are two specific points located on the major axis of the ellipse. These points have a unique geometrical property. For any point on the ellipse, the sum of the distances to the two foci is constant. This characteristic is crucial in defining the shape and properties of the ellipse. In our exercise, the foci are given at

• (0, 4)
• (0, -4)

This tells us that the ellipse's major axis is aligned with the y-axis. Understanding the foci positions helps determine other key parameters of the ellipse.

Vertices of an Ellipse

The vertices of an ellipse are the points where the ellipse is widest or longest. For a vertical ellipse, like the one in our example, the vertices lie on the y-axis. The given vertices for this ellipse are:

• (0, 5)
• (0, -5)

These points indicate that the semi-major axis extends 5 units from the center of the ellipse at the origin (0, 0). Knowing the vertices is essential, as they assist in finding other critical ellipse parameters.

Standard Form of an Ellipse

The standard form equation of an ellipse allows us to understand its geometry and dimensions. For our vertical ellipse, the standard form is: \[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\]Here, 'a' represents the length of the semi-major axis, and 'b' is the length of the semi-minor axis. In our ellipse, because the major axis is vertical, the 'a' value is associated with the 'y' component in the equation. Substituting the known values, the equation becomes:\[\frac{x^2}{9} + \frac{y^2}{25} = 1\]The standard equation expresses a clear and structured way to represent the ellipse's dimensions and orientation.

Semi-Major Axis

The semi-major axis is the longest radius from the center to the perimeter of an ellipse. It is essentially half of the major axis, which passes through both the center and the ellipse's two farthest points (vertices). In the given example, the length of the semi-major axis, 'a', is determined by the vertices.

• Vertex at (0, 5) implies a semi-major axis length of 5.
• This is confirmed by the relationship \(a = 5\).

The semi-major axis helps define the shape and positioning of an ellipse, with 'a' playing a crucial role in the standard form equation.

Semi-Minor Axis

The semi-minor axis of an ellipse is the shortest radius from the center to the ellipse's perimeter. It is a key component perpendicular to the semi-major axis. The semi-minor axis length, represented as 'b', can be calculated when we know 'a' and the distance 'c' from the center to each focus. In the current problem, 'b' resolves through:

• The relationship \(c^2 = a^2 - b^2\)
• Given \(c = 4\) and \(a = 5\), then \(b^2 = 9\)

This result means that the semi-minor axis is 3 units long. Understanding 'b' is vital in completing the ellipse's geometry, especially within the standard form equation.