Problem 3 - Algebra 2

Question

Comets or other objects that pass by Earth or the Sun only once and never return may follow hyperbolic paths. Suppose a comet's path can be modeled by a branch of the hyperbola with equation \(\frac{y^{2}}{225}-\) \(\frac{x^{2}}{400}=1 .\) Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola. Then graph the hyperbola.

Step-by-Step Solution

Answer

Vertices: (0, 15), (0, -15); Foci: (0, 25), (0, -25); Asymptotes: \(y = \pm \frac{3}{4}x\).

Step 1: Identify the Hyperbola Equation

The given equation is \(\frac{y^2}{225} - \frac{x^2}{400} = 1\). This is in the standard form for a vertical hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Therefore, we identify \(a^2 = 225\) and \(b^2 = 400\).

Step 2: Calculate the Vertices

For a vertical hyperbola centered at the origin, the vertices are located at \((0, \pm a)\). Since \(a^2 = 225\), we find \(a = 15\). Hence, the vertices are \((0, 15)\) and \((0, -15)\).

Step 3: Calculate the Foci

The distance of the foci from the center is given by \(c\), where \(c^2 = a^2 + b^2\). Substitute \(a^2 = 225\) and \(b^2 = 400\) to find \(c^2 = 225 + 400 = 625\), thus \(c = 25\). Therefore, the foci are at \((0, 25)\) and \((0, -25)\).

Step 4: Write the Equation of the Asymptotes

For the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{a}{b}x\). Here, \(a = 15\) and \(b = 20\), so the asymptotes are \(y = \pm \frac{15}{20}x\) or \(y = \pm \frac{3}{4}x\).

Step 5: Graph the Hyperbola

To graph the hyperbola, plot the vertices \((0, 15)\) and \((0, -15)\), and the asymptotic lines \(y = \frac{3}{4}x\) and \(y = -\frac{3}{4}x\) as guides. The hyperbola will open upwards and downwards passing through the vertices, and approaching the asymptotes as \(x\) moves away from the center along its branch.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are specific points on the hyperbola itself, and they are crucial in determining its size and shape. For a vertical hyperbola, like the one provided in the exercise, the vertices can be found along the y-axis.
The general formula to find the vertices of a vertical hyperbola centered at the origin is

• (0, \(a\)) and (0, -\(a\))

where \(a\) is the distance from the center to each vertex.

Given the equation \(\frac{y^2}{225} - \frac{x^2}{400} = 1\), you can extract \(a^2 = 225\). Solving for \(a\) gives \(a = 15\), so the vertices in this case are at the points \( (0, 15)\) and \( (0, -15)\).

These vertices define the width of the hyperbola along its vertical axis and help indicate the direction in which the branches of the hyperbola open. Since this is a vertical hyperbola, the branches open upwards and downwards.

Foci of Hyperbola

The foci are another critical component of a hyperbola, found inside each branch of the curve. The distance from the center to each focus is denoted by \(c\).
The relationship between \(a\), \(b\), and \(c\) for a hyperbola

• centered at the origin is given by \(c^2 = a^2 + b^2\).

To find the foci of the given hyperbola, first calculate \(c\).

We've already found \(a^2 = 225\) and \(b^2 = 400\). Plugging these into the formula gives

• \(c^2 = 225 + 400 = 625\)

resulting in \(c = 25\).

This results in the foci being located at

• (0, 25) and (0, -25).

These points assist in understanding how stretched the hyperbola is along the y-axis and are also critical for constructing an accurate graph of the hyperbola.

Asymptotes of Hyperbola

Asymptotes of a hyperbola are linear guides that indicate the directions that the branches of the hyperbola tend to grow towards. They play a crucial role in sketching a hyperbola, especially when trying to visualize it without detailed plotting.
For a hyperbola, represented by the equation

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

where the hyperbola is vertical, the equations of the asymptotes can be found using \(y = \pm \frac{a}{b}x\).

In the exercise, replace \(a = 15\) and \(b = 20\), into the formula, which results in the asymptotic equations:

• \(y = \frac{3}{4}x\)
• \(y = -\frac{3}{4}x\).

These asymptotes inform the user that the branches will grow closer and closer to the lines \(y = \pm \frac{3}{4}x\) as one moves further from the center of the hyperbola. Having these lines plotted on the graph will help ensure the hyperbola is drawn accurately and proportionately.

Equation of Hyperbola

The equation of a hyperbola in its standard form provides a great deal of information about the hyperbola's orientation and key features.
For a vertical hyperbola centered at the origin, the equation is typically presented as

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

In our case from the exercise, this translates into the equation \(\frac{y^2}{225} - \frac{x^2}{400} = 1\).

This equation tells us several things:

• The vertices can be found along the y-axis because the variable \(y^2\) appears first, confirming that this is a vertical hyperbola.
• The values of \(a^2 = 225\) and \(b^2 = 400\) allow for calculating both vertices and asymptotes.
• The calculation of the foci’s distance from the center uses \(c^2 = a^2 + b^2\).

Having this equation enables a clearer and structured approach to finding out all other parameters like the foci, vertices, and asymptotes. It serves as the foundation for not only mathematical solutions but also for graphing the hyperbola. Furthermore, understanding the equation aids in comparing different hyperbolas easily.