Problem 51 - Precalculus: Functions and Graphs

Question

Find an equation for the set of points in an xy-plane such that the difference of the distances from \(F\) and \(F\) is \(k\) $$F(13,0), \quad F(-13,0) ; \quad k=24$$

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{144} - \frac{y^2}{25} = 1 \).

Step 1: Identify the Problem Type

The exercise describes finding an equation for a set of points where the difference in the distances to two fixed points (foci) is constant. This is the definition of a hyperbola.

Step 2: Standard Form of a Hyperbola Equation

For a hyperbola centered at the origin, with foci at \( (c, 0) \) and \( (-c, 0) \), the equation is \\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]\where \(c^2 = a^2 + b^2\).

Step 3: Determine the Values of c, a, and b

Given the foci are \( (13,0) \) and \( (-13,0) \), the value of \( c \) is 13. The constant difference in distances, \( k \), is 24. This implies \( 2a = 24 \), so \( a = 12 \).

Step 4: Calculate b Using the Relationship Between a, b, and c

Use the equation \( c^2 = a^2 + b^2 \) to find \( b^2 \): \c^2 = 169, \ a^2 = 144 \\[ 169 = 144 + b^2 \]\\[ b^2 = 25 \]\Thus, \( b = 5 \).

Step 5: Write the Equation of the Hyperbola

Now that we have \( a^2 = 144 \) and \( b^2 = 25 \,\) we can write the equation: \\[ \frac{x^2}{144} - \frac{y^2}{25} = 1 \].

Key Concepts

Conic sections

Conic sections are curves obtaining by intersecting a plane with a cone. These intersections yield different shapes, each with unique properties. The main types of conic sections include:

• Circle: A set of points equidistant from a center point.
• Ellipse: An oval shape where the sum of distances from any point on the ellipse to two foci is constant.
• Parabola: A curve where any point is equidistant from a point known as the focus and a line called the directrix.
• Hyperbola: This is characterized by open curves where the difference of distances from any point on the curves to two fixed points, called foci, is constant.

Each conic section carries distinct algebraic equations and characteristics. In precalculus, working with these sections helps build a foundational understanding of geometric shapes and their algebraic representations. The hyperbola, which results from the plane intersecting both nappes of the cone, is particularly interesting for its role in many scientific contexts, including physics and astronomy.

Foci of a hyperbola

The foci of a hyperbola are crucial to its definition. They are two fixed points located symmetrically on either side of the hyperbola’s center along the transverse axis. Let's explore why they're important and how they relate to the hyperbola's properties.
In a hyperbola, the absolute difference in distances from any point on the curve to these two foci remains constant. This unique property shapes the hyperbolic structure and is central to its reflective characteristics. Unlike an ellipse where the sum of distances to the foci is constant, the hyperbola relies on the difference.

The distance between the center and each focus is signified by the variable \(c\). Hence, if the foci are located at \((c, 0)\) and \((-c,0)\), the separation measurement helps in deriving its equation. Understanding the role of the foci is pivotal in graphing hyperbolas, as it informs us about their orientation and stretch.

Difference of distances

The defining feature of a hyperbola is that the difference of the distances from any point on its curve to the two fixed points, called foci, is constant. This characteristic means:

• If you take any point on one branch of the hyperbola and measure distances to each focus, then subtract one result from the other, the value remains the same for any point you choose on that hyperbola.
• This constant difference, denoted by \(k\), helps precisely define the shape and size of the hyperbola.

In practice, if a problem provides the difference is \(k = 24\), for instance, it gives direct insight into the values needed to calculate the hyperbola's "a" value, since \(2a = k\). Knowing this property is essential when transforming the hyperbola from a geometric definition into an algebraic equation, a key step in precalculus.

Equation derivation in precalculus

Deriving the equation of a hyperbola involves a deep understanding of its symmetry and defining features. In precalculus, this skill is developed to connect the graphic and algebraic representations effectively. Here's how this process usually unfolds:
First, identify key characteristics such as center, foci, and the constant difference \(k\). Establish whether your hyperbola's orientation is horizontal or vertical. For hyperbolas centered at the origin and horizontally oriented (like our example), the equation takes the form:

\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

Next, use the relationship \(c^2 = a^2 + b^2\) to determine the missing components once city's coordinates (\((c,0)\) and \((-c,0)\)) and \(k\) are known.

Here's where the practical derivation happens:

• Given, \(c = 13\) and \(k = 24\), we recognize \(2a = 24\) giving \(a = 12\).
• Substitute into \(c^2 = a^2 + b^2\):
  \[ 169 = 144 + b^2 \]
  leading to \(b^2 = 25\) (thus, \(b = 5\)).
• Finally, articulate the hyperbola's equation using these a and b values.

This step-by-step breakdown grounds students in both geometric understanding and algebraic manipulation, ensuring a robust grasp of hyperbolas in precalculus.