Problem 34 - Precalculus with Limits

Question

In Exercises 29-34, find the standard form of the equation of the hyperbola with the given characteristics and center at the origin. Foci: \((\pm10, 0); \quad\) asymptotes: \(y=\pm\frac{3}{4}\)

Step-by-Step Solution

Answer

The standard equation of the hyperbola is \( \frac{9x^2}{1600} - \frac{4y^2}{225} = 1 \)

Step 1: Identify 'c' from the given foci

The foci of the hyperbola are given as \((\pm10, 0)\), from this we identify that \(c = 10\). This comes from the definition of the foci in a hyperbola.

Step 2: Identify the ratio \(\frac{b}{a}\) from the asymptote

The slope of the asymptotes is given by \(y=\pm\frac{3}{4}\). Since the slope of the asymptotes in a hyperbola is given by \(\pm \frac{b}{a}\), we can equate this to the given slope. So, \(\frac{b}{a}= \frac{3}{4}\)

Step 3: Solve for 'a' and 'b'

We now have two equations, \(\frac{b}{a} = \frac{3}{4}\) and \(c^2=a^2-b^2\) with \(c=10\). We can solve these two equations to find the values of 'a' and 'b'. Solving, \[a = \frac{4c}{3} = \frac{4*10}{3} = \frac{40}{3}\] and \[b = \frac{3c}{4} = \frac{3*10}{4} = \frac{30}{4} = 7.5\]

Step 4: Write the standard equation

Now that we have 'a' and 'b', we can write the equation of the hyperbola in standard form. \[\left(\frac{x^2}{(\frac{40}{3})^2}\right) - \left(\frac{y^2}{7.5^2}\right) = 1\] Simplifying, we have \[\frac{9x^2}{1600} - \frac{4y^2}{225} = 1\]

Key Concepts

Standard Form of Hyperbola

A hyperbola is a type of conic section that comes with its own standard equation. For a hyperbola centered at the origin, the standard form of the equation is:

• For a hyperbola that opens horizontally: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
• For a hyperbola that opens vertically: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]

In these equations, \(a\) and \(b\) represent the distances related to the hyperbola's axes. Specifically, \(a\) is the distance from the center to each vertex on the transverse axis, and \(b\) relates to the direction of the asymptotes. The equation choice (whether horizontal or vertical) depends on which term is expressed positively.

Hyperbola Asymptotes

Asymptotes are lines that the hyperbola approaches, but never actually meets. They play a crucial role in determining the shape of the hyperbola. The slopes of these asymptotes are a defining feature. For a hyperbola centered at the origin, the asymptotes are:

• For horizontal hyperbolas: \[ y = \pm \frac{b}{a}x \]
• For vertical hyperbolas: \[ y = \pm \frac{a}{b}x \]

By interpreting the slopes from the asymptotes' equations, we can find the relative lengths of \(a\) and \(b\). In our given problem, the slope is \(\frac{3}{4}\), leading to \(\frac{b}{a} = \frac{3}{4}\). Asymptotes effectively "guide" the hyperbola, illustrating its broadest direction as it extends infinitely.

Hyperbola Characteristics

Understanding the characteristics of a hyperbola involves identifying its parts:

• Center: The middle point from which the hyperbola is defined. In this case, it is the origin point \((0,0)\).
• Vertices: These are the closest points on each branch of the hyperbola to its center. They lie along the transverse axis at a distance \(a\) from the center.
• Foci: These points lie within each branch of the hyperbola and are located at a distance \(c\) from the center. Given as \((\pm10, 0)\), this is critical for determining \(c^2 = a^2 + b^2\).
• Axes: The transverse axis is the line segment between the vertices, and the conjugate axis is perpendicular to it. These help determine the orientation of the hyperbola.

Each characteristic ties back to the standard form equation and provides a complete picture of the hyperbola's size and shape.