Problem 28 - Algebra 2

Question

Graph each equation. $$ 16 x^{2}-20 y^{2}=560 $$

Step-by-Step Solution

Answer

The given equation represents a hyperbola with its center at the origin (0, 0). The vertices are at (\(-\sqrt{35}\), 0) and (\(\sqrt{35}\), 0), the foci are at (\(-\sqrt{63}\), 0) and (\(\sqrt{63}\), 0), and the asymptotes are \(y = ±\frac{\sqrt{28}}{\sqrt{35}}x\). The graph will open to the right and left.

Step 1: Rearrange the equation into standard form

The given equation is \(16x^2 - 20y^2 = 560\). Divide the entire equation by 560 to put it into standard form. You get \(\frac{x^2}{35} - \frac{y^2}{28} = 1\).

Step 2: Identify the characteristics of the hyperbola

The standard form of the equation of a hyperbola is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) where (h, k) is the center of the hyperbola. In our rearranged equation, h = 0, k = 0, a = sqrt(35), and b = sqrt(28). Thus, the center of the hyperbola is at the origin (0,0).

Step 3: Find the foci

The formula for the foci is \( f = \sqrt{a^2 + b^2}\). This gives \(f = \sqrt{35 + 28} = \sqrt{63}\). So the foci are at \((-\sqrt{63}, 0)\) and \((\sqrt{63}, 0)\).

Step 4: Identify the asymptotes

The equation for the asymptotes of a horizontal hyperbola is \(y = ± \frac{b}{a}x\). Plugging in our a and b values results in \( y = ±\frac{\sqrt{28}}{\sqrt{35}}x\).

Step 5: Draw the graph

Place the center at the origin. Draw the asymptotes as diagonal lines passing through the origin. The vertices are at \((-\sqrt{35}, 0)\) and \((\sqrt{35}, 0)\). The foci are at \((-\sqrt{63}, 0)\) and \((\sqrt{63}, 0)\). Since the x term is positive, the hyperbola opens to the right and left. The foci and vertices lie on the x axis.

Key Concepts

Hyperbola Standard Form

To understand a hyperbola, it's essential to start by expressing its equation in standard form. This makes it easier to identify its characteristics. For a hyperbola that opens horizontally, the standard form is:

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

Here:

• \( (h, k) \) is the center of the hyperbola.
• \( a^2 \) and \( b^2 \) are the denominators of the fractional terms on the right side that help define the hyperbola's shape and size.

For our exercise, we rearrange the given equation \(16x^2 - 20y^2 = 560\) to standard form:

• Divide the entire equation by 560: \( \frac{x^2}{35} - \frac{y^2}{28} = 1 \)
• This reveals that \( h = 0, k = 0, \) \( a = \sqrt{35} \), and \( b = \sqrt{28} \).

This step sets the groundwork for interpreting and graphing a hyperbola.

Foci of Hyperbola

The foci of a hyperbola are crucial points that define its shape and direction. Unlike ellipses where the sum of distances to the foci is constant, in hyperbolas, the difference is constant.To find the foci, use this formula:

• \[ f = \sqrt{a^2 + b^2} \]

Where \( a^2 \) and \( b^2 \) are the squared values from the standard form equation. For our problem:

• \( a^2 = 35 \) and \( b^2 = 28 \), thus \( f = \sqrt{35 + 28} = \sqrt{63} \).

This means you place the foci at:

• \( (-\sqrt{63}, 0) \) and \( (\sqrt{63}, 0) \).

Remember, the foci determine the location of the hyperbola's two branches, helping it open sideways along the x-axis in this case.

Hyperbola Asymptotes

Asymptotes are the diagonal lines that a hyperbola approaches but never touches. They provide a boundary within which the hyperbola exists.For horizontal hyperbolas, the equations of the asymptotes are:

• \[ y = \pm \frac{b}{a}x \]

Given \( a = \sqrt{35} \) and \( b = \sqrt{28} \), plug these into the formula:

• \[ y = \pm \frac{\sqrt{28}}{\sqrt{35}}x \]

The asymptotes can be graphically represented as lines passing through the origin, cutting through the coordinate plane diagonally. These lines aid in sketching the hyperbola, knowing it will hug but not intersect these lines. They help visualize how wide the branches of the hyperbola will spread.

Vertices of Hyperbola

The vertices of a hyperbola are the closest points to the center. They mark the tips of the hyperbola’s two open ends. For horizontal hyperbolas:

• Vertices are located along the x-axis, at \((h \pm a, k)\).

In our case, with the center at \((0,0)\) and \(a = \sqrt{35}\):

• The vertices are found at \((-\sqrt{35}, 0)\) and \((\sqrt{35}, 0)\).

These vertices are critical for sketching the hyperbola's structure accurately as they guide the curve's opening and direction. The line segment between these vertices is often called the transverse axis, and it stretches along the hyperbola’s direction of opening. Understanding vertices is integral in capturing the hyperbola's distinct shape in graph form.