Problem 24 - Thomas Calculus

Question

Exercises \(17-24\) give equations for ellipses. Put each equation in standard form. Then sketch the ellipse. Include the foci in your sketch. $$ 169 x^{2}+25 y^{2}=4225 $$

Step-by-Step Solution

Answer

The standard form is \( \frac{x^2}{25} + \frac{y^2}{169} = 1 \). Foci are at (0,12) and (0,-12).

Step 1: Identify the General Form

The given equation is in the form \( Ax^2 + By^2 = C \), which is \( 169x^2 + 25y^2 = 4225 \). This indicates an equation of an ellipse centered at the origin.

Step 2: Divide by the Constant Term

To transform the equation into standard form, divide every term by \( 4225 \) so that the right side equals 1: \[ \frac{169x^2}{4225} + \frac{25y^2}{4225} = 1 \].

Step 3: Simplify Each Term

Simplify each fraction: \( \frac{169}{4225} = \frac{169}{25^2} \) and \( \frac{25}{4225} = \frac{25}{25^2} \). This yields: \[ \frac{x^2}{25} + \frac{y^2}{169} = 1 \].

Step 4: Identify the Components

Now the equation \( \frac{x^2}{25} + \frac{y^2}{169} = 1 \) is in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a^2 = 25 \), \( b^2 = 169 \). Thus, \( a = 5 \) and \( b = 13 \). This indicates that the semi-major axis runs along the y-axis.

Step 5: Determine Foci

For ellipses, the foci are located at \( (0,\pm c) \). Calculate \( c \) using \( c^2 = b^2 - a^2 \): \( c^2 = 169 - 25 = 144 \) so \( c = 12 \). The foci are at \( (0, 12) \) and \( (0, -12) \).

Step 6: Sketch the Ellipse

Draw a vertical ellipse with center at (0,0), semi-major axis length 13 along the y-axis, and semi-minor axis length 5 along the x-axis. Include foci at (0,12) and (0,-12).

Key Concepts

Standard Form of Ellipse

An ellipse is a set of points such that the sum of the distances from two fixed points (called foci) is constant. The standard form of an ellipse is written based on its orientation, either horizontal or vertical. In the standard form, an ellipse equation can appear as

• Horizontal: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where \( a > b \), and the x-axis is the major axis.
• Vertical: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \), where \( b > a \), and the y-axis is the major axis.

The key to converting the general form of an ellipse equation into standard form is to make the right side equal to 1. You achieve this by dividing all terms by the constant on the right side. Once simplified, the fractions correspond to the squares of the semi-major and semi-minor axes, \( a^2 \) and \( b^2 \).

In the exercise given, initially, the equation \(169x^2 + 25y^2 = 4225\) was in the form \( Ax^2 + By^2 = C \). By dividing every term by 4225, we arrive at the standard form: \[ \frac{x^2}{25} + \frac{y^2}{169} = 1 \] Here, \( a = 5 \) and \( b = 13 \), indicating it's a vertically oriented ellipse since \( b > a \).

Foci of Ellipse

Foci (plural of focus) are essential parts of an ellipse, and their locations give the ellipse its unique shape. For a vertically oriented ellipse, the foci are positioned along the y-axis at a distance \( c \) from the center, where \( c \) can be calculated using the formula:

• \( c^2 = b^2 - a^2 \)

This is derived from the fact that in an ellipse, the distance from the center to a focus is less than the semi-major axis length.

For the exercise's standard ellipse equation \( \frac{x^2}{25} + \frac{y^2}{169} = 1 \), we have \( a^2 = 25 \) and \( b^2 = 169 \). Thus, to find \( c \): \[ c^2 = 169 - 25 = 144 \] This implies \( c = 12 \). The foci are therefore located at points \( (0, 12) \) and \( (0, -12) \) from the center at the origin (0,0).

The foci play a crucial role in defining the ellipse's shape, since any point on the ellipse will have the sum of distances to the foci being constant.

Ellipse Sketching

Drawing an ellipse is easier once you have calculated the semi-major and semi-minor axes and located the foci. A clear and precise sketch helps visualize and grasp the geometric properties of the ellipse. Here is how you can sketch the ellipse from our example:First, determine the center of the ellipse, which in this case is the origin (0,0). The semi-major axis, as dictated by the larger denominator \( b^2 = 169 \), is 13 units along the y-axis. Thus, the ellipse extends 13 units up and down from the center.

Next, note the semi-minor axis, determined by \( a^2 = 25 \), which is 5 units along the x-axis, extending the ellipse 5 units to the left and right from the center.Finally, place the foci at \( (0, 12) \) and \( (0, -12) \) on the y-axis. With these points, draw a smooth curve connecting the ends of the axes, creating an elongated circle.

This sketch allows you to clearly see how the foci add the ellipse's defining feature, being nearer than the endpoints of the longest axis and at its true center.