Problem 35

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: $(\pm 1,0),$ asymptotes: $y=\pm 5 x$

Step-by-Step Solution

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Answer

The equation of the hyperbola is $x^2 - \frac{y^2}{25} = 1$.

1Step 1: Identify the type of hyperbola

Given that the vertices of the hyperbola are at $(\pm 1, 0)$, this means the hyperbola is centered at the origin $(0, 0)$ with a horizontal transverse axis. This indicates that the equation of the hyperbola will be of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

2Step 2: Use the vertices to find $a$

Since the vertices are at $(\pm 1, 0)$, the distance from the center to each vertex is $a = 1$. Thus, $a^2 = 1$.

3Step 3: Use the asymptotes to find $b$

The given asymptotes are $y = \pm 5x$. For a hyperbola with a horizontal transverse axis, the equations of the asymptotes are $y = \pm \frac{b}{a} x$. Since $ \frac{b}{a} = 5 $ and $a = 1$, it follows that $b = 5$. Hence, $b^2 = 25$.

4Step 4: Write the equation of the hyperbola

Now plug the values of $a^2$ and $b^2$ into the standard form of the hyperbola's equation: $\frac{x^2}{1} - \frac{y^2}{25} = 1$. The equation simplifies to: $x^2 - \frac{y^2}{25} = 1$.

Key Concepts

Vertices of HyperbolaAsymptotes of HyperbolaStandard Form of HyperbolaTransverse AxisHorizontal Transverse Axis

Vertices of Hyperbola

In the study of hyperbolas, vertices play a critical role in defining the shape and position of the curve. They are the points where the hyperbola intersects its transverse axis. For a hyperbola centered at the origin and with a horizontal transverse axis, the vertices will be along the x-axis. For example, if the vertices are

(1, 0) and (-1, 0),

this implies that the distance from the center to each vertex is 1 unit. This distance is denoted by the letter $a$. Thus, in this case, $a = 1$ because the vertices are at $x = \pm 1$. Vertices mark the width of the hyperbola at its narrowest point, and they are pivotal in determining the standard equation of the hyperbola.

Asymptotes of Hyperbola

Asymptotes are diagonal lines that the hyperbola approaches but never meets as it extends towards infinity. For hyperbolas with a horizontal transverse axis, asymptotes serve as guides for the curve. They can be expressed as lines intersecting at the hyperbola's center. In this case,
the asymptotes are:

$y = 5x$
$y = -5x$.

These lines provide crucial information about the hyperbola's end behavior. For horizontal hyperbolas, the slope $\pm \frac{b}{a}$ of the asymptotes helps in identifying $b$, where $b$ is the distance from the center parallel to the asymptotes. Given $\frac{b}{a} = 5$ and that $a = 1$, we deduce $b = 5$. Asymptotes help us understand how steep the hyperbola is.

Standard Form of Hyperbola

When dealing with hyperbolas, their equations can be expressed in a standard form, which is straightforward to work with. For horizontal hyperbolas centered at the origin, the standard form of the equation is:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1.\]This formula contains two expressions:

$ \frac{x^2}{a^2} $ indicating the horizontal distance related to the vertices,
$ \frac{y^2}{b^2} $ related to the vertical extension governed by the slopes of the asymptotes.

In our example, substituting $a = 1$ and $b = 5$, the equation becomes\[x^2 - \frac{y^2}{25} = 1.\]This provides a clear picture of how the hyperbola opens and spreads out from its center.

Transverse Axis

The transverse axis of a hyperbola is a critical feature as it runs through the center and both vertices, defining the direction in which the hyperbola opens. For the given exercise, it's horizontal, meaning the hyperbola's vertices stretch along the x-axis. The length of the transverse axis is computed as $2a$. In this example, where the vertices are