Problem 25 - Algebra and Trigonometry

Question

In Exercises \(25-36,\) find the standard form of the equation of each ellipse satisfying the given conditions. Foci: \((-5,0),(5,0) ;\) vertices: \((-8,0),(8,0)\)

Step-by-Step Solution

Answer

The standard form of the equation of the ellipse is \(\frac{x^2}{64} + \frac{y^2}{39} = 1\)

Step 1: Identify the center, 'a', and 'c'

Given are the foci \((-5,0)\) and \( (5,0)\), the center of the ellipse is the midpoint of these points, which is \((0,0)\). The total distance from one focus to the other is 10, hence \(c = 5\). The vertices are \((-8,0)\) and \( (8,0)\), therefore the total distance from one vertex to the other is 16, thus \(a = 8\).

Step 2: Calculate 'b'

Knowing that \(a^2 = b^2 + c^2\), we can substitute the found values of 'a' and 'c' in this equation to find 'b'. \(b^2 = a^2 - c^2 = 8^2 - 5^2 = 39\), therefore \(b = \sqrt{39}\)

Step 3: Write the standard form of the ellipse equation

Having the values for 'a', 'b' and the center coordinates, substitute these into the standard form equation of the ellipse. The equation then becomes \[\frac{x^2}{8^2} + \frac{y^2}{(\sqrt{39})^2} = 1\], simplifying gives \[\frac{x^2}{64} + \frac{y^2}{39} = 1\]

Key Concepts

Foci of an Ellipse

An ellipse has two special points known as the foci (singular: focus). These points are crucial for defining the geometry of the ellipse. The sum of the distances from any point on the ellipse to each focus is constant. This property gives the ellipse its stretched circle-like shape. In the provided exercise, the foci are at

• e of the foci is at (-5, 0)
• and the other is at (5, 0)

You determine the center of the ellipse by finding the midpoint of the segment joining these two points. Therefore, the center in this case is at the origin, (0, 0). The distance between the foci (known as the distance, 2c) provides a measure of how "stretched" the ellipse is. Here, the total distance is 10, thus each focus is 5 units from the center (c = 5). Understanding the role of foci aids in visualizing how the ellipse behaves in terms of its shape and orientation.

Vertices of an Ellipse

The vertices of an ellipse are the points where the ellipse is longest and lie on the same line as the foci. These points indicate the end of the major axis. In a horizontal ellipse like the one in this exercise, the vertices extend out from the center along the x-axis. Provided are the vertices at

• (-8, 0)
• (8, 0)

The total distance from one vertex to the other is known as the major axis and for our ellipse is 16. As such, the value of 'a', representing half this length, is 8. Recognizing the position of the vertices helps sketch the main structure of the ellipse and understand its scale. They are always "a" units away from the center in either direction along the major axis.

Ellipse Center

The center of an ellipse is a point that serves as a reference in positioning the rest of its features, such as vertices and foci. For an ellipse with a horizontal major axis, this is also the midpoint between the foci and the midpoint between the vertices. Determining the center allows us to symmetrically place the ellipse’s other components around it. In the given exercise, this point is simple to determine, being right at (0, 0). By calculating the midpoint of the foci (-5, 0) and (5, 0) , we confirm that the ellipse is centered at the origin. Understanding the center’s location is vital for graphing the ellipse and applying other transformations.

Standard Form Equation of an Ellipse

The equation for an ellipse standard form depends on its orientation and size. For a horizontally oriented ellipse centered at (0, 0), the standard form is given by\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]Here, 'a' is the semi-major axis and 'b' is the semi-minor axis. Plugging the provided values into the equation, we have\[ a = 8 \quad and \quad c = 5 \]We use the relationship \( a^2 = b^2 + c^2 \) to solve for 'b'. Therefore,\( b^2 = 64 - 25 = 39 \),which gives us \( b = \sqrt{39} \).With these, the equation becomes\[ \frac{x^2}{64} + \frac{y^2}{39} = 1 \].This formula portrays the balance between the major and minor axes respective to the distances from the center. Understanding this form allows you to not only sketch the ellipse but also solve problems involving elliptical shapes in geometry.