Problem 43 - Algebra 2

Question

Find the vertices and the asymptotes of each hyperbola. $$ 16 x^{2}-25 y^{2}=400 $$

Step-by-Step Solution

Answer

The vertices of the given hyperbola are at (5,0) and (-5,0) and the equations of the asymptotes are \(y = \frac{4}{5}x\) and \(y = -\frac{4}{5}x\).

Step 1: Writing in Standard Form

Express the given equation \(16x^{2} - 25y^{2} = 400\) in standard form of a hyperbola equation. The standard form of a hyperbola equation that opens left and right is \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\). Divide the entire equation by 400 to get the standard form: \(\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1\).

Step 2: Determine a and b

The values under x and y in the equation are \(a^{2}\) and \(b^{2}\) respectively. Hence, \(a = 5\) and \(b = 4\).

Step 3: Calculate c

We find c by using the equation \(c = \sqrt{a^{2} + b^{2}}\). Substituting the values of a and b, we get \(c = \sqrt{5^{2} + 4^{2}} = \sqrt{41}\).

Step 4: Identify Vertices

The vertices will occur at \(\pm a, 0\). Thus, the vertices of this hyperbola are at \((5, 0)\) and \((-5, 0)\).

Step 5: Identify Asymptotes

The equations for the asymptotes in a standard hyperbola equation that opens left and right are given by \(y = \pm \frac{b}{a}x\). Thus, for this hyperbola, the asymptotes are \(y = \pm \frac{4}{5}x\).

Key Concepts

Vertices of a Hyperbola

Vertices of a hyperbola are crucial points that define its shape and orientation in the coordinate plane. They are the points where the hyperbola intersects its principal axis. In the context of the given equation, the vertices can be found once the equation is in its standard form.
To locate the vertices for a hyperbola that opens left and right, we use the formula for the standard form:

• For horizontal hyperbolas: i.e., \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at \((\pm a, 0)\).
• After calculating \(a\) from the equation,\(a^2 = 25\), the value of \(a\) is 5. This tells us that the hyperbola's vertices are at \((5, 0)\) and \((-5, 0)\).

The vertices indicate the points of closest approach between the branches of the hyperbola. Understanding the location of vertices helps in sketching the hyperbola correctly.

Asymptotes of a Hyperbola

Asymptotes of a hyperbola give a guideline for how the two branches of the hyperbola extend infinitely on the coordinate plane. Unlike vertices, asymptotes are lines the hyperbola approaches but never touches.
To find the equations of the asymptotes for a hyperbola along the standard equation:

• For horizontal hyperbolas where the equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \),the asymptotes are given by: \[ y = \pm \frac{b}{a}x \]
  1. Plugging in \(b = 4\) and \(a = 5\) from the standard form transformation, the asymptotes are represented by the equations \(y = \pm \frac{4}{5}x\).

Asymptotes play an important role because they act as the invisible boundary lines that guide the hyperbola's shape and direction.

Standard Form of Hyperbola

The standard form of a hyperbola equation is essential for analyzing its properties such as vertices and asymptotes. In hyperbolas, the standard form equation is different based on the orientation of the hyperbola:

• For hyperbolas opening left and right (horizontal):\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
• For hyperbolas opening up and down (vertical):\( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \)

In the exercise, the given equation \(16x^{2} - 25y^{2} = 400\) was converted to its standard form by dividing all terms by 400:

1. This yields \(\frac{x^{2}}{25} - \frac{y^{2}}{16} = 1\), indicating a horizontally oriented hyperbola.
2. The expression shows clearly what the denominators represent, with \(a^2 = 25\) and \(b^2 = 16\).

By rewriting the equation in this standard form, it becomes straightforward to identify important features like vertices and asymptotes, contributing to our overall understanding of the hyperbola's geometry.