Problem 13
Question
Graph each hyperbola. $$ \frac{x^{2}}{4}-\frac{y^{2}}{9}=1 $$
Step-by-Step Solution
Verified Answer
Graph the hyperbola with center at (0,0), vertices at (2,0) and (-2,0), co-vertices at (0,3) and (0,-3), and asymptotes y = ±(3/2)x.
1Step 1: Identify the Standard Form of the Hyperbola
The given equation \( \frac{x^{2}}{4} - \frac{y^{2}}{9} = 1 \) is in the standard form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), where \( a^2 = 4 \) and \( b^2 = 9 \). This indicates a hyperbola centered at the origin \( (0, 0) \) that opens horizontally.
2Step 2: Calculate the Values of a and b
To find \( a \) and \( b \), take the square root of \( a^2 \) and \( b^2 \). Thus, \( a = \sqrt{4} = 2 \) and \( b = \sqrt{9} = 3 \). This information helps determine the distances from the center to the vertices and co-vertices.
3Step 3: Determine the Vertices
The vertices of a horizontal hyperbola \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) are located at \( (\pm a, 0) \). Therefore, the vertices of this hyperbola are located at \( (2, 0) \) and \( (-2, 0) \).
4Step 4: Determine the Co-vertices
The co-vertices are located \( b \) units from the center along the y-axis. Thus, they are at \( (0, \pm 3) \), which are \( (0, 3) \) and \( (0, -3) \).
5Step 5: Sketch the Asymptotes
The asymptotes for this hyperbola are given by the equations \( y = \pm \frac{b}{a}x \). Substituting the values: \( \pm \frac{3}{2}x \), which means the equations are \( y = \frac{3}{2}x \) and \( y = -\frac{3}{2}x \). These lines help guide the sketching of the hyperbola.
6Step 6: Graph the Hyperbola
Plot the center at \( (0,0) \), the vertices at \( (2,0) \) and \( (-2,0) \), and the co-vertices at \( (0,3) \) and \( (0,-3) \). Use the asymptotes as guidelines to sketch the two branches of the hyperbola. The graph consists of two separate curves opening horizontally away from each other.
Key Concepts
Understanding the Hyperbola's Standard FormVertices and Co-vertices: Key Points of the HyperbolaImportance of Hyperbola AsymptotesGraphing Techniques for Hyperbolas
Understanding the Hyperbola's Standard Form
The equation of the hyperbola given in this exercise is \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \). This is known as the standard form of a hyperbola.
In general, the standard form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, the hyperbola opens horizontally. If the signs in the standard form are flipped, like \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the hyperbola opens vertically.
The constants \( a^2 \) and \( b^2 \) play crucial roles in understanding the hyperbola's dimensions. In our equation, \( a^2 = 4 \) and \( b^2 = 9 \). By comparing these with the general form, we identify the orientation and shape of the hyperbola. This provides valuable insight into how you should expect the graph to appear.
In general, the standard form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, the hyperbola opens horizontally. If the signs in the standard form are flipped, like \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the hyperbola opens vertically.
The constants \( a^2 \) and \( b^2 \) play crucial roles in understanding the hyperbola's dimensions. In our equation, \( a^2 = 4 \) and \( b^2 = 9 \). By comparing these with the general form, we identify the orientation and shape of the hyperbola. This provides valuable insight into how you should expect the graph to appear.
Vertices and Co-vertices: Key Points of the Hyperbola
Vertices and co-vertices are vital points that help us sketch the hyperbola accurately.
For a hyperbola with the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), and centered at the origin, the vertices are situated at \((\pm a, 0)\). This gives us the primary axis along which the hyperbola stretches.
Substituting \( a = 2 \) (since \( \sqrt{4} = 2 \)), we find the vertices at \((2, 0)\) and \((-2, 0)\).
Co-vertices are positioned along the other axis, the y-axis for our horizontal hyperbola. They are found at \((0, \pm b)\). Here \( b = 3 \) (since \( \sqrt{9} = 3 \)), so the co-vertices are \((0, 3)\) and \((0, -3)\).
These points not only define the extent of the hyperbola but also aid in drawing precise asymptotic lines.
For a hyperbola with the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), and centered at the origin, the vertices are situated at \((\pm a, 0)\). This gives us the primary axis along which the hyperbola stretches.
Substituting \( a = 2 \) (since \( \sqrt{4} = 2 \)), we find the vertices at \((2, 0)\) and \((-2, 0)\).
Co-vertices are positioned along the other axis, the y-axis for our horizontal hyperbola. They are found at \((0, \pm b)\). Here \( b = 3 \) (since \( \sqrt{9} = 3 \)), so the co-vertices are \((0, 3)\) and \((0, -3)\).
These points not only define the extent of the hyperbola but also aid in drawing precise asymptotic lines.
Importance of Hyperbola Asymptotes
Asymptotes are lines that the hyperbola approaches but never actually touches. For the hyperbola \( \frac{x^2}{4} - \frac{y^2}{9} = 1 \), the asymptotes serve as diagonal guidelines.
The equations for asymptotes in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \( y = \pm \frac{b}{a}x \). This means we calculate the slope of each asymptote. With our values \( b = 3 \) and \( a = 2 \), the asymptotes become \( y = \frac{3}{2}x \) and \( y = -\frac{3}{2}x \).
They cross at the hyperbola's center \((0,0)\), forming a sort of 'X' that guides the hyperbola's branch pathways. Remember, though, the graph itself never crosses these lines, no matter how the curves extend.
The equations for asymptotes in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \( y = \pm \frac{b}{a}x \). This means we calculate the slope of each asymptote. With our values \( b = 3 \) and \( a = 2 \), the asymptotes become \( y = \frac{3}{2}x \) and \( y = -\frac{3}{2}x \).
They cross at the hyperbola's center \((0,0)\), forming a sort of 'X' that guides the hyperbola's branch pathways. Remember, though, the graph itself never crosses these lines, no matter how the curves extend.
Graphing Techniques for Hyperbolas
Graphing a hyperbola involves a combination of plotting key features and following specific guidelines. First, identify the center of the hyperbola, which in our exercise is at \((0, 0)\). This is your starting plot point.
Then, place the vertices \((2, 0)\) and \((-2, 0)\), and co-vertices \((0, 3)\) and \((0, -3)\). These points frame your hyperbola, establishing the major and minor axes.
Next, sketch the asymptotes \( y = \frac{3}{2}x \) and \( y = -\frac{3}{2}x \). Use a ruler or a straight edge for straight and precise lines. These will help in drawing the two symmetrical branches of the hyperbola accurately.
Finally, draw smooth curves starting from one vertex, hugging close to the asymptotes as they extend in opposite directions and repeat for the second branch. This results in a hyperbola that visually captures the solution's requirements.
Then, place the vertices \((2, 0)\) and \((-2, 0)\), and co-vertices \((0, 3)\) and \((0, -3)\). These points frame your hyperbola, establishing the major and minor axes.
Next, sketch the asymptotes \( y = \frac{3}{2}x \) and \( y = -\frac{3}{2}x \). Use a ruler or a straight edge for straight and precise lines. These will help in drawing the two symmetrical branches of the hyperbola accurately.
Finally, draw smooth curves starting from one vertex, hugging close to the asymptotes as they extend in opposite directions and repeat for the second branch. This results in a hyperbola that visually captures the solution's requirements.
Other exercises in this chapter
Problem 12
Graph each inequality. $$ y
View solution Problem 12
Solve each nonlinear system of equations. $$ \left\\{\begin{array}{l} y=x+1 \\ x^{2}-y^{2}=1 \end{array}\right. $$
View solution Problem 13
The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. See Examples 1 through 4. $$ y=x^{2}+10 x+20 $$
View solution Problem 13
Graph each inequality. $$ \frac{x^{2}}{4}+\frac{y^{2}}{9} \leq 1 $$
View solution