Problem 48 - Algebra and Trigonometry with Analytic Geometry

Question

Find an equation for the set of points in an \(x y\)-plane such that the difference of the distances from \(F\) and \(F^{\prime}\) is \(k\). $$F(0,17), \quad F^{\prime}(0,-17) ; \quad k=30$$

Step-by-Step Solution

Answer

The equation is \(\frac{y^2}{225} - \frac{x^2}{64} = 1\).

Step 1: Understanding the Problem

We need to find the equation for the set of points (locus) where the difference of distances from two fixed points (foci) is constant. This describes a hyperbola.

Step 2: Identifying the Foci and Constant

The foci of the hyperbola are given as points \(F(0, 17)\) and \(F^{\prime}(0, -17)\). The constant difference \(k\) is given as \(30\).

Step 3: Recognizing the Hyperbola Equation

For a hyperbola, the difference between distances from any point \((x, y)\) on the hyperbola to the two foci is constant. The general form of a vertical hyperbola with center at the origin is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

Step 4: Calculating the Vertices

The distance between the foci is \(34\) (because \(17 - (-17) = 34\)). We use this to find \(2c\), where \(c\) is the distance from the center to each focus. We have \(2c = 34\), giving \(c = 17\).

Step 5: Finding 'a' and 'b'

For a hyperbola, the relationship \(|2a| = k\) holds. Here, \(a = \frac{30}{2} = 15\). With \(c = 17\), use the relationship \(c^2 = a^2 + b^2\) to find \(b^2\): \(17^2 = 15^2 + b^2\) which gives \(b^2 = 64\).

Step 6: Writing the Hyperbola Equation

Substitute \(a^2 = 225\) and \(b^2 = 64\) into the hyperbola equation: \(\frac{y^2}{225} - \frac{x^2}{64} = 1\), which is the equation of the hyperbola satisfying the given condition.

Key Concepts

Distance

In the given problem, distance plays a crucial role in defining a hyperbola. A hyperbola is formed by the set of points where the difference in distances to two fixed points, known as the foci, is constant. This constant difference is a key characteristic that distinguishes hyperbolas from other conic sections like ellipses and circles. The formula \(d_1 - d_2 = k\) is used, where \(d_1\) and \(d_2\) are the distances from any point on the hyperbola to the foci, and \(k\) represents the constant difference.

• For the given hyperbola, this difference, denoted by \(k\), is 30.
• The points are equidistant in a way that this difference remains constant across the curve.

Understanding how this difference forms the basis of the hyperbola helps in comprehending its geometric properties and equation.

Foci

Foci are central to the definition of a hyperbola. Specifically, these are the two fixed points used to derive the set of all points (the curve) satisfying the distance condition. For the hyperbola in the exercise, the foci are given as \(F(0, 17)\) and \(F'(0, -17)\).

• The foci lie on the y-axis because their x-coordinates are both zero.
• The role of the foci is essential as they are used to measure the constant difference in distances that define the hyperbola.

By focusing on the coordinates and arrangement of these foci, we determine that this is a vertical hyperbola since its foci are aligned along the vertical axis.

Locus

The term locus refers to the set of points that satisfy specific conditions. In the context of a hyperbola, it is the collection of points whose distances from the foci have a constant difference. The problem involves finding this set of points, which is expressed as the hyperbola.

• The points on the locus are described by the vertex, foci, and the equation of the curved path they form.
• Understanding the relationship of the locus to the foci and distance helps visualize what the hyperbola looks like in a coordinate plane.

The locus is fundamentally the "shape" of the hyperbola, offering insight into how its points behave uniformly under the distance condition.

Vertical Hyperbola

A vertical hyperbola is identified by its orientation; it opens up and down along the y-axis. This occurs when the foci of the hyperbola are vertically aligned, as seen in the given points \(F(0, 17)\) and \(F'(0, -17)\). The standard equation for such a hyperbola is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).

• The difference \(d_1 - d_2 = k\) remains constant as you move along the curve vertically.
• Calculations show \(a = 15\) and \(b^2 = 64\), leading to the specific equation \(rac{y^2}{225} - rac{x^2}{64} = 1\).

This orientation and the inclusion of specific constants \(a\) and \(b\) define the size and shape of the hyperbola, reflecting its opening along the vertical axis.