Problem 20 - Algebra and Trigonometry

Question

use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ 9 y^{2}-x^{2}=1 $$

Step-by-Step Solution

Answer

The vertices of the hyperbola are at (0, 3) and (0, -3), the foci are at (0, sqrt(10)) and (0, -sqrt(10)), the equations of the asymptotes are y = ±3x and y = ∓3x, and the graph of the hyperbola opens upwards and downwards, bounded by the asymptotes.

Step 1: Rewrite the Equation

Write the equation \( 9y^2 - x^2 = 1 \) in standard form \(\frac{(y - h)^{2}}{a^{2}} - \frac{(x - k)^{2}}{b^{2}} = 1\), where (h, k) is the center of the hyperbola. Here, h = k = 0, a^2 = 9 and b^2 = 1. So the standard form is \(\frac{y^2}{9} - \frac{x^2}{1} = 1\)

Step 2: Identify the Vertices, Asymptotes, and Foci

Since this is a vertical hyperbola, the vertices are (0, a) and (0, -a), or (0, 3) and (0, -3). The distance from the center to a focus is \(c\), where \(c^2 = a^2 + b^2\). Therefore, \(c = \sqrt{9 + 1} = \sqrt{10}\), so the foci are (0, +sqrt(10)) and (0, -sqrt(10)). The equations of the asymptotes for a vertical hyperbola are y = ±a/b * x + h and y = ± -a/b * x + k. Thus the equations of the asymptotes are y = ±3x and y = ∓3x.

Step 3: Graph the Hyperbola

Plot the center, vertices, and foci on the coordinate plane. Draw the asymptotes, which are the diagonal lines y = ±3x. Sketch the two branches of the hyperbola, which open upwards and downwards, and are bounded by the asymptotes.

Key Concepts

Standard Form of a Hyperbola

Understanding the standard form of a hyperbola is crucial for graphing and analyzing its properties. The standard form equation for a hyperbola that opens vertically, like in our exercise, is expressed as:
\frac{(y - h)^{2}}{a^{2}} - \frac{(x - k)^{2}}{b^{2}} = 1
Where (h, k) are the coordinates of the hyperbola's center. The variables 'a' and 'b' represent the distances from the center to the vertices and to the points on the conjugate axis, respectively. To rewrite the given equation
9y^2 - x^2 = 1
in standard form, we follow these steps:

• Rearrange the terms to separate x's and y's.
• Divide by the leading coefficients to achieve the '1' on the right side of the equation.
• Adjust the terms to reveal the subtraction like in the standard form.

As a result, the transformed standard form equation is
\(\frac{y^2}{9} - \frac{x^2}{1} = 1\).
This representation helps us to identify all the other significant elements of a hyperbola, such as vertices, asymptotes, and foci.

Vertices of a Hyperbola

The vertices of a hyperbola are key points that lie on the hyperbola at the ends of its primary axis. For the given hyperbola with the standard form equation
\(\frac{y^2}{9} - \frac{x^2}{1} = 1\),
the vertices are located at a distance 'a' above and below the center along the y-axis, since it opens vertically. In our exercise, with
\(a^2 = 9\)
and thus
\(a = 3\),
the vertices are determined by
\((h, k \text{±} a)\)
which translates to (0, 3) and (0, -3) when the center is at the origin (0,0). It's important for students to note that identifying the vertices is one of the first steps to graphing the hyperbola as it gives a clear indication of the shape and size of the curve.

Asymptotes of a Hyperbola

Asymptotes are imaginary lines that a hyperbola approaches but never touches. They provide a guideline for sketching the shape of the hyperbola. For vertical hyperbolas like in our exercise, the equations of the asymptotes are derived from the standard form as
\(y = \text{±} \frac{a}{b} x + h\)
and
\(y = \text{±} -\frac{a}{b} x + k\).
Using the values from the standard form
\(\frac{y^2}{9} - \frac{x^2}{1} = 1\),
where
\(a = 3\),
\(b = 1\),
and the center is at (0,0), we get the equations
\(y = \text{±} 3x\).
These two lines intersect at the hyperbola's center and demonstrate how to draw the branches of the hyperbola accurately.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points located inside each branch of the hyperbola that play an essential role in its definition. The distance of each focus from the center is derived using the equation
\(c^2 = a^2 + b^2\).
In our exercise's hyperbola
\(\frac{y^2}{9} - \frac{x^2}{1} = 1\),
with
\(a^2 = 9\)
and
\(b^2 = 1\),
we calculate
\(c\) as
\(c = \sqrt{9 + 1} = \sqrt{10}\).
Therefore, the foci are located at (0, +\sqrt{10}) and (0, -\sqrt{10}), which are
\(\sqrt{10}\)
units above and below the center along the y-axis for our vertically oriented hyperbola. When graphing the hyperbola, students need to ensure that they accurately plot the foci as they will affect the curvature of the branches.