Problem 27 - Precalculus: Functions and Graphs

Question

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Vertices \(V \pm 4,0), \quad\) passing through \((8,2)\)

Step-by-Step Solution

Answer

\(\frac{x^2}{16} - \frac{3y^2}{4} = 1\)

Step 1: Identify the Hyperbola Type

Since the vertices are given as \((\pm 4, 0)\), the hyperbola is oriented along the x-axis. Therefore, the standard form of the hyperbola's equation is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Step 2: Determine 'a' from Vertices

The distance from the center of the hyperbola to a vertex is the value of 'a'. Since the vertex is at (\(\pm 4, 0)\), we have \(a = 4\).

Step 3: Use the Standard Equation

Substitute \(a = 4\) into the standard form. This gives: \[ \frac{x^2}{4^2} - \frac{y^2}{b^2} = 1 \] which simplifies to \[ \frac{x^2}{16} - \frac{y^2}{b^2} = 1 \].

Step 4: Plug in the Given Point

Substitute the point \((8, 2)\) into the hyperbola equation to find \(b^2\). Substitute \(x = 8\) and \(y = 2\): \[ \frac{8^2}{16} - \frac{2^2}{b^2} = 1 \].

Step 5: Solve for 'b^2'

Calculate \(\frac{8^2}{16} - \frac{2^2}{b^2} = 1\). Simplify to get \(4 - \frac{4}{b^2} = 1\). Solve for \(b^2\): \(3 = \frac{4}{b^2}\), hence \(b^2 = \frac{4}{3}\).

Step 6: Write the Final Equation

Substitute \(b^2 = \frac{4}{3}\) back into the hyperbola equation to get: \[ \frac{x^2}{16} - \frac{y^2}{\frac{4}{3}} = 1 \]. This can also be written as \[ \frac{x^2}{16} - \frac{3y^2}{4} = 1 \] by multiplying through by 3.

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double-napped cone. This may result in different shapes depending on the angle and direction of intersection.
These include:

• Circles: Formed when the intersection is perpendicular to the axis of the cone.
• Ellipses: Occur when the intersection is at an angle, but the plane does not go through the base of the cone.
• Parabolas: Result from intersecting the cone parallel to its side.
• Hyperbolas: Created when the plane intersects both nappes of the cone.

Hyperbolas are distinct as they consist of two disconnected curves. They're defined by their asymptotes, axes, vertices, and foci, making them a fascinating part of conic sections.

Equation of Hyperbola

The equation of a hyperbola depends on its orientation and position. For a hyperbola centered at the origin, its equation can be written in two standard forms based on its axis of orientation:

• Along the x-axis: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
• Along the y-axis: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

The values of \(a^2\) and \(b^2\) determine the stretches along the axes. In general, remember that \(a^2\) always corresponds to the axis aligned with the hyperbola's opening direction, while the sign in the equation determines whether it opens horizontally or vertically.
Furthermore, these equations are equipped to calculate other properties like vertices and foci.

Vertices of Hyperbola

The vertices of a hyperbola are the points where the hyperbola intersects its principal axis. These are essential in defining the shape and orientation of the hyperbola.
For a hyperbola centered at the origin and aligned horizontally, the vertices are:

• \((\pm a, 0)\)

The distance from the center to each vertex is equal to \(a\). For our particular problem, given that vertices are at \((\pm4, 0)\), this shows that \(a = 4\).
These vertices assist in pinpointing the length of the transverse axis, which is critical to the overall representation of the hyperbola, making them a vital component of hyperbolic geometry.

Standard Form of Hyperbola

A hyperbola's standard form provides insight into its geometric properties. It fundamentally depends on whether the hyperbola is horizontally or vertically oriented.
For a horizontal hyperbola centered at the origin, the standard form is:

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

This configuration emphasizes the formula's characteristics, dictating the hyperbola's stretches along the x and y axes.By plugging in given properties, such as vertices and points the hyperbola passes through, you can derive specific values of \(a^2\) and \(b^2\). This allows the precise drawing and analysis of the hyperbola within the plane. Understanding this standard form creates an essential toolkit for students solving problems related to hyperbolic curves.

Center of Hyperbola

The center of a hyperbola is the midpoint between its vertices and its foci. It's a crucial reference point for positioning all key hyperbolic components. When a hyperbola is centered at the origin

• \((0, 0)\)

It simplifies calculations and graphing. The center acts as the cornerstone from which other elements like vertices and asymptotes emanate. From the center, hyperbolas are symmetrically organized around their principal axes. For problems like the given one, knowing the center allows for identifying equations quickly, which revolve around this focal point in the coordinate system.