Problem 55 - Algebra 2

Question

Write the equation of each hyperbola in standard form. Sketch the graph. $$ 16 x^{2}-10 y^{2}=160 $$

Step-by-Step Solution

Answer

The standard form of the given equation is \(x^2/10 - y^2/16 = 1\) and it opens along the x-axis with vertices at \(\pm \sqrt{10}\)

Step 1: Simplify the equation

Start by dividing the equation by the constant on the right side, in this case, 160, to set the equation equal to 1. This simplifies to \(x^2/10 - y^2/16 = 1\).

Step 2: Rearrange into standard form

The simplified equation is nearly in standard form. However, the 'y' term must be made positive, we simply rearrange the terms: \(x^2/10 - y^2/16 = 1\) becomes \(x^2/10 - y^2/16 = 1\). Nothing changes as this is simply rearrangement.

Step 3: Graph the hyperbola

To graph, first draw the two axes. The square root of 10 and 16 gives the distances from the origin along the x and y axes respectively, which are known as the semi-major and semi-minor axes. As 'x' is positive in this equation, the hyperbola opens along the x-axis. Draw lines through these points as guides to sketch the hyperbola curves. The vertex of each hyperbola curve is on the x-axis at \(\pm \sqrt{10}\), and the asymptotes of the hyperbola pass through the origin and the points \(\pm \sqrt{10}\), \(\pm \sqrt{16}\). The traces intersect the x-axis at \(\pm \sqrt{10}\), indicating the vertices, and the y-axis at \(\pm \sqrt{16}\) which are the focal points. Hence, we have the curves opening along the x-axis with vertices at \(\pm \sqrt{10}\).

Key Concepts

Standard Form of Hyperbola

When dealing with hyperbolas, it is crucial to express their equations in a standard form. The standard form allows us to easily identify key features of the hyperbola such as the center, vertices, and axes. For a hyperbola centered at the origin, the standard form is either

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

These forms tell us whether the hyperbola opens horizontally or vertically.
In the exercise, the equation \( 16x^2 - 10y^2 = 160 \) was given. To rewrite it in standard form, divide every term by 160:

• This gives us \( \frac{x^2}{10} - \frac{y^2}{16} = 1 \).

Now, this equation fits the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), indicating a horizontally opening hyperbola centered at the origin with a transverse axis along the x-axis.

Graphing Hyperbolas

Once a hyperbola is written in standard form, graphing it becomes much more straightforward. In our example, the standard form equation is \( \frac{x^2}{10} - \frac{y^2}{16} = 1 \). Since the 'x' term is positive, we know the hyperbola opens horizontally.
To graph this hyperbola:

• Calculate the vertices by finding \( a = \sqrt{10} \). The vertices are located at \( \pm \sqrt{10} \) on the x-axis.
• The semi-minor axis, found using \( b = \sqrt{16} = 4 \), helps determine the shape of the hyperbola.
• Draw these distances on the corresponding axes to outline the box that guides your sketch.
• Draw the asymptotes of the hyperbola. These are lines that pass through the origin with slopes given by \( \pm b/a = \pm \sqrt{16}/\sqrt{10} \).

With these guides, smoothly sketch the hyperbola, ensuring the curves approach but never touch the asymptotes.

Equation Simplification

Before dealing with the details of graphing or identifying features of a hyperbola, we often start with simplification to get an equation into a usable form. Simplification generally involves isolating terms or factoring, making computational work easier.
For hyperbolas, the first step is typically setting the equation equal to 1 by dividing through by the constant. In our exercise:

• The original equation \( 16x^2 - 10y^2 = 160 \) was divided by 160.
• This results in \( \frac{x^2}{10} - \frac{y^2}{16} = 1 \), a much more workable form.

By simplifying, each term is transformed into a fraction that reveals the relation between the x and y variables straightforwardly. This simplification is essential, as it makes it far easier to identify the hyperbola's orientation and to graph it accurately.