Problem 65
Question
Find an equation for each hyperbola. $$\text { Asymptotes } y=\pm \frac{3}{5} x, y \text { -intercepts }(0, \pm 3)$$
Step-by-Step Solution
Verified Answer
The equation is \(\frac{y^2}{9} - \frac{x^2}{25} = 1\).
1Step 1: Analyze the asymptotes
The given asymptotes are \( y = \pm \frac{3}{5}x \). This implies that the slopes of the asymptotes are \( \pm \frac{3}{5} \). In the standard equation of a hyperbola, the asymptotes for a vertical hyperbola (centered at the origin) are \( y = \pm \frac{a}{b}x \). Hence, we infer that \( \frac{a}{b} = \frac{3}{5} \).
2Step 2: Utilize the intercepts
We are given that the \( y \)-intercepts are \( (0, \pm 3) \). This means that the hyperbola intersects the \( y \)-axis at these points. For a vertical hyperbola, the points where it crosses the \( y \)-axis are \( (0, \pm a) \), thus \( a = 3 \).
3Step 3: Relate asymptotes and axes lengths
From the relationship \( \frac{a}{b} = \frac{3}{5} \) and \( a = 3 \), we substitute to find \( b \):\[\frac{3}{b} = \frac{3}{5} \Rightarrow b = 5.\]
4Step 4: Write the equation of the hyperbola
With \( a = 3 \) and \( b = 5 \), the equation of a vertical hyperbola centered at the origin is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Substitute \( a = 3 \) and \( b = 5 \) to get: \[\frac{y^2}{9} - \frac{x^2}{25} = 1.\]
Key Concepts
AsymptotesY-interceptsVertical hyperbolaEquation of a hyperbola
Asymptotes
In the context of hyperbolas, asymptotes play a crucial role in understanding their geometry. Asymptotes are straight lines that a hyperbola approaches but never actually meets. They act as boundaries that guide the hyperbola’s shape. For our hyperbola, the asymptotes are given by the equations \( y = \pm \frac{3}{5}x \). This indicates that the slopes of these lines are \( \pm \frac{3}{5} \).
In hyperbolas, particularly those centered at the origin, these slopes are connected to the relationship \( \frac{a}{b} \). This specific relationship gives insight into the ratio of the lengths of the hyperbola's axes.
In hyperbolas, particularly those centered at the origin, these slopes are connected to the relationship \( \frac{a}{b} \). This specific relationship gives insight into the ratio of the lengths of the hyperbola's axes.
- A positive slope implies the hyperbola will rise to the right.
- A negative slope implies it falls to the right.
Y-intercepts
Finding the \( y \)-intercepts is another key part of understanding how a hyperbola behaves. The \( y \)-intercepts of our hyperbola are at \((0, \pm 3)\). These points signify where the hyperbola crosses the \( y \)-axis.
For vertical hyperbolas, like the one in this problem, these intercepts translate to meaningful information about the hyperbola's axes. Specifically, the intercepts tell us the value of \( a \), which is the distance from the center to the vertex along the \( y \)-axis. In our case, \( a = 3 \).
Why does this matter?
For vertical hyperbolas, like the one in this problem, these intercepts translate to meaningful information about the hyperbola's axes. Specifically, the intercepts tell us the value of \( a \), which is the distance from the center to the vertex along the \( y \)-axis. In our case, \( a = 3 \).
Why does this matter?
- The value of \( a \) helps to complete the equation of the hyperbola by informing us about the vertical stretch.
- The value of \( a \) is used to determine the hyperbola's steepness and shape.
Vertical hyperbola
Vertical hyperbolas are a specific type of hyperbola that open up and down, as opposed to horizontally. These are characterized by having their transverse axis (the segment containing the vertices) along the \( y \)-axis. In any standard vertical hyperbola equation, the \( y^2 \) term appears first, indicating the orientation.
Formally, the standard form of the equation can be recognized thus:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]This structure contrasts with horizontal hyperbolas where the \( x^2 \) term appears first.
Formally, the standard form of the equation can be recognized thus:\[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\]This structure contrasts with horizontal hyperbolas where the \( x^2 \) term appears first.
- In our example, the arrangement of terms confirms the hyperbola opens vertically.
- The term \( y^2 \) dictates that the hyperbola is aligned along the vertical \( y \)-axis.
Equation of a hyperbola
An equation of a hyperbola is a significant part of its definition and helps specify its exact shape and orientation. In this specific instance, the hyperbola is characterized by the equation:\[\frac{y^2}{9} - \frac{x^2}{25} = 1\]The equation reveals crucial aspects about the hyperbola.
The \( y^2 \) being divided by 9 (or \( a^2 \)) indicates that the hyperbola is vertical with \( a = 3 \). Similarly, \( x^2 \) being divided by 25 (or \( b^2 \)) shows \( b = 5 \).
The \( y^2 \) being divided by 9 (or \( a^2 \)) indicates that the hyperbola is vertical with \( a = 3 \). Similarly, \( x^2 \) being divided by 25 (or \( b^2 \)) shows \( b = 5 \).
- The fraction \( \frac{y^2}{9} \) signifies relation to the \( y \)-axis, impacting the width of the branches.
- The fraction \( \frac{x^2}{25} \) relates to a lateral distance from the center to the asymptotes.
Other exercises in this chapter
Problem 64
Each equation in Exercises defines a parabola. Without actually graphing. match each equation with the appropriate description. $$x=-(y-2)^{2}-4$$ A. Vertex \((
View solution Problem 65
Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}+y^{2} &=10 \\ 2 x^{2}-y^{2} &=17 \end{aligned}$$
View solution Problem 65
For the graph of \((x-h)^{2}=4 c(y-k)\) in what quadrant is the vertex for each condition? (a) \(h0\) (c) \(h>0, k0, k>0\)
View solution Problem 66
Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}+y^{2} &=4 \\ 2 x^{2}-3 y^{2} &=-12 \end{aligned}$$
View solution