Problem 37 - Precalculus with Limits

Question

In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((4, 1), (4, 9); \quad\) foci: \((4, 0), (4, 10)\)

Step-by-Step Solution

Answer

The required standard equation of the hyperbola is \((\frac{(y-5)^2}{16}) - (\frac{(x-4)^2}{9})= 1\)

Step 1: Identify the center of the hyperbola

The center of the hyperbola is the midpoint between either the vertices or the foci. In this case, the center can be found by averaging the coordinates of the vertices. Thus, the center of the hyperbola is \((4, \frac{1+9}{2})\), simplifying to (4,5).

Step 2: Identify the value of a and c

The value of \(a\) is the distance from the center to a vertex. Since a vertex is at (4,1) and the center is at (4,5), \(a\) is 4 units.\nThe value of \(c\) is the distance from the center to a focus. Since a focus is at (4,0) and the center is at (4,5) \(c\) is 5 units.

Step 3: Calculate the value of b

Next, find the value of \(b\), this can be found using the relationship \(c^2 = a^2 + b^2\). Re-write the equation as \(b^2 = c^2 - a^2 = 5^2 - 4^2\), which is 9. Hence, \(b = \sqrt{9} = 3\)

Step 4: Write the equation of the hyperbola

Since the vertices are vertically aligned, the standard equation of a hyperbola is \((\frac{(y-k)^2}{a^2})- (\frac{(x-h)^2}{b^2})=1. Where (h,k) represents the center, a represents the distance from the center to the vertices, and b represents the distance from the center to the co-vertices. Replace h,k,a,b into the standard equation to get \((\frac{(y-5)^2}{4^2}) - (\frac{(x-4)^2}{3^2})=1

Key Concepts

Understanding Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. These include ellipses, parabolas, circles, and hyperbolas. Each shape has unique characteristics and equations that describe them.
Hyperbolas emerge when the plane intersects the cone at an angle parallel to its axis. They are made up of two separate curves called the branches of the hyperbola.
The study of conic sections is fundamental in mathematics as these figures can be used to model real-world phenomena ranging from planetary orbits to radio wave signals.
Understanding hyperbolas involves recognizing their geometrical properties and translating them into equations that describe their shape and orientation.

Standard Form Equation of a Hyperbola

The standard form equation of a hyperbola depends on its orientation—whether it's vertical or horizontal. For a vertical hyperbola like in the given exercise, the standard form is:

• \(\left(\frac{(y - k)^2}{a^2}\right) - \left(\frac{(x - h)^2}{b^2}\right) = 1\)

Here,

• \((h, k)\) is the center of the hyperbola.
• \(a\) is the distance from the center to each vertex.
• \(b\) is determined from \(b^2 = c^2 - a^2\).

For horizontal hyperbolas, the equation would be:

• \(\left(\frac{(x - h)^2}{a^2}\right) - \left(\frac{(y - k)^2}{b^2}\right) = 1\)

Using the standard form makes it easier to identify and understand the conic's properties such as vertices, foci, and asymptotes.

Applying the Midpoint Formula

The midpoint formula is a mathematical method used to find the center point between two given points on a plane. It is especially useful in determining the center of figures like hyperbolas and other conics.
The formula is given by:

• \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)

In hyperbolas, the center is found by applying the midpoint formula either to the vertices or foci coordinates. In our exercise, the vertices were used. By applying this formula to the coordinates

• \((4, 1) \) and \((4, 9)\),
• the center is calculated as \((4, 5)\).

Knowing the center allows precise determination of the hyperbola’s equation.

Using the Distance Formula

The distance formula helps in calculating the exact distance between two points in a coordinate plane. It is vital in determining key features of hyperbolas, such as the distance from the center to vertices and foci.
The formula is:

• \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

To find the distance for this exercise, we calculated:

• Distance from the center \((4, 5)\) to a vertex \((4, 1)\),
• which provided \(a = 4\).
• Distance from the center to a focus \((4, 0)\),
• which gave \(c = 5\).

These measurements fulfill the relationship \(c^2 = a^2 + b^2\), essential for finding \(b^2\) and completing the standard hyperbola equation. The distance formula ensures these dimensions are mathematically accurate.