Problem 43 - Precalculus: Functions and Graphs

Question

Exer. \(41-44:\) Find an equation for the set of points in an Xy-plane such that the sum of the distances from \(F\) and \(F\) is \(k\) $$F(0,15), \quad F(0,-15) ; \quad k=34$$

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{64} + \frac{y^2}{289} = 1 \).

Step 1: Understand the Problem

We need to find an equation for the set of points where the sum of the distances from two fixed points \( F_1(0,15) \) and \( F_2(0,-15) \) is equal to a constant \( k = 34 \). This is the definition of an ellipse.

Step 2: Recall the Standard Form of an Ellipse

An ellipse with foci at \( (0,c) \) and \( (0,-c) \) has the equation: \[ \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \] where \( a \geq b \) and the sum of distances from the foci to any point on the ellipse is \( 2a \).

Step 3: Relate Given Parameters to the Standard Form

In our problem, the sum of distances is given as \( k = 34 \), which means \( 2a = 34 \), therefore \( a = 17 \). The foci \( c \) are located at \( 15 \), thus \( c = 15 \).

Step 4: Use the Relationship Between \( a \), \( b \), and \( c \) in Ellipse

The relationship is given as \( c^2 = a^2 - b^2 \). Substitute \( a = 17 \) and \( c = 15 \) into this equation: \( 15^2 = 17^2 - b^2 \).

Step 5: Solve for \( b^2 \)

Calculate \( 17^2 = 289 \) and \( 15^2 = 225 \). Hence, \[ 225 = 289 - b^2 \]. Solving this gives \[ b^2 = 64 \].

Step 6: Write the Equation of the Ellipse

Now plug \( a = 17 \) and \( b^2 = 64 \) into the standard form equation: \[ \frac{x^2}{64} + \frac{y^2}{289} = 1 \]. This is the equation of the ellipse.

Key Concepts

Foci of an Ellipse

In an elliptical shape, the two fixed points are known as the **foci**. For a clearer picture, imagine an ellipse having focus points on each of its two sides.
For the ellipse defined in the problem, the foci are at the coordinates

• \((0,15)\) and \((0,-15)\) .

These coordinates mark where the foci are located along the y-axis.
The foci are crucial in defining the specific shape of the ellipse. They are used in the distance sum definition of an ellipse, tying into how an ellipse is mathematically mapped.
The properties of an ellipse primarily revolve around its foci, as they help determine the ellipse's orientation and dimensions.

Standard Form of an Ellipse

The standard equation of an ellipse provides a mathematical way to understand its structure. This equation is given as:
\[\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\]
Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively. For this problem:

• \(a\) is
  • the semi-major axis; since it's greater than or equal to \(b\).

The equation derives from the distance between the foci and any point on the ellipse. In our case, this ellipse has a specific configuration:

• \(b^2 = 64\)
• \(a^2 = 289\)

By plugging these values into the standard form, you get:\[\frac{x^2}{64} + \frac{y^2}{289} = 1\]
This tells us how the ellipse is shaped in the xy-plane based on its axes lengths.

Distance Sum Definition of Ellipse

The fundamental property of an ellipse rests on the distance sum definition. If you take any point on the ellipse and measure the distance to each of the two foci, the sum of these distances remains constant.
For this exercise, let’s denote this constant sum as \(k\).

• The relationship is given by:
  • \(2a = k\), where \(a\) is the semi-major axis.

Given that \(k = 34\), we find that \(2a = 34\); thus, \(a = 17\).
This definition is pivotal for understanding how an ellipse can be formed and how it behaves.
It shows that no matter which direction the ellipse stretches, the total distance from any point on its boundary to the foci is always \(34\), providing a consistent geometric balance.