Problem 40
Question
Identify whether equation, when graphed, will be a parabola, circle, ellipse, or hyperbola. Sketch the graph of equation. If a parabola, label the vertex. If a circle, label the center and note the radius. If an ellipse, label the center. If a hyperbola, label the \(x\) - or \(y\) -intercepts. \(\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\)
Step-by-Step Solution
Verified Answer
The equation is a hyperbola, with x-intercepts at (4, 0) and (-4, 0).
1Step 1: Recognize standard equation forms
The given equation is \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \). It resembles the standard form of a hyperbola, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where the \(x^2\) term is positive and comes first. This indicates it is a horizontal hyperbola.
2Step 2: Identify the elements of the hyperbola
Compare the given equation \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \) with \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 16 \) and \( b^2 = 4 \), so \( a = 4 \) and \( b = 2 \). For the hyperbola, the center is at (0,0), since the equation does not have \(x\) or \(y\) constants.
3Step 3: Finding the intercepts
For a hyperbola, the \(x\)-intercepts can be found by setting \(y = 0\) in the equation. Substituting \(y = 0\) gives \( \frac{x^2}{16} = 1 \), which simplifies to \( x^2 = 16 \). Therefore, \(x = \pm 4\). This means the hyperbola has \(x\)-intercepts at (+4, 0) and (-4, 0).
4Step 4: Sketch the graph
Draw the coordinate plane and plot the center of the hyperbola at (0, 0). Then sketch the transverse axis along the \(x\)-axis, showing the intercepts at (4, 0) and (-4, 0). Remember that a hyperbola with a horizontal transverse axis opens left and right through these intercepts. Roughly sketch the curves of the hyperbola.
Key Concepts
HyperbolaEquation of a HyperbolaGraphing HyperbolasX-intercepts of a HyperbolaStandard Form of Conics
Hyperbola
A hyperbola is a type of conic section that is created when a plane intersects both nappes of a double cone. Unlike ellipses and circles, which are closed curves, hyperbolas open outward indefinitely. The basic structure of a hyperbola consists of two disconnected curves called branches, which mirror each other across the hyperbola's center.
A hyperbola can be recognized by its characteristic shape and equation, which includes a subtraction of squared terms. Understanding its properties, such as asymptotes and intercepts, is crucial for sketching its graph accurately.
A hyperbola can be recognized by its characteristic shape and equation, which includes a subtraction of squared terms. Understanding its properties, such as asymptotes and intercepts, is crucial for sketching its graph accurately.
Equation of a Hyperbola
The standard equation for a hyperbola in the Cartesian coordinate system is written differently depending on whether the transverse axis is horizontal or vertical. A horizontal hyperbola has the following form:
In these equations, \(a\) and \(b\) represent distances from the center of the hyperbola to the vertices and to the co-vertices, respectively. This configuration allows us to determine the directions in which the hyperbola opens. Also, it aids in finding key elements like intercepts and vertices.
- Horizontal: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
- Vertical: \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
In these equations, \(a\) and \(b\) represent distances from the center of the hyperbola to the vertices and to the co-vertices, respectively. This configuration allows us to determine the directions in which the hyperbola opens. Also, it aids in finding key elements like intercepts and vertices.
Graphing Hyperbolas
Drawing a hyperbola requires identifying its critical components, such as the center, vertices, and intercepts. First, locate the center of the hyperbola. For equations in their standard forms, the center is at the origin \((0,0)\) unless shifted by additional constants.
Next, plot the vertices. For a horizontal hyperbola like \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \), the vertices are located at \((-a, 0)\) and \((a, 0)\), here \((4,0)\) and \((-4,0)\). Graph the asymptotes by using the slopes \(\frac{b}{a}\) and \(-\frac{b}{a}\); these lines guide the opening of the hyperbola branches around them, but are not part of the hyperbola itself. Finally, sketch the branches emanating from the vertices along the direction of the asymptotes.
Next, plot the vertices. For a horizontal hyperbola like \( \frac{x^2}{16} - \frac{y^2}{4} = 1 \), the vertices are located at \((-a, 0)\) and \((a, 0)\), here \((4,0)\) and \((-4,0)\). Graph the asymptotes by using the slopes \(\frac{b}{a}\) and \(-\frac{b}{a}\); these lines guide the opening of the hyperbola branches around them, but are not part of the hyperbola itself. Finally, sketch the branches emanating from the vertices along the direction of the asymptotes.
X-intercepts of a Hyperbola
The intercepts of a hyperbola are points where it crosses the axes. Finding the x-intercepts involves setting \(y = 0\) in the equation and solving for \(x\).
For the equation \(\frac{x^2}{16} - \frac{y^2}{4} = 1\), substituting \(y = 0\) gives \(\frac{x^2}{16} = 1\). Solving this, we find \(x^2 = 16\), which gives the solutions \(x = \pm 4\). These calculations show that the hyperbola crosses the x-axis at \((4, 0)\) and \((-4, 0)\). These intercepts are critical for plotting the hyperbola accurately, as they define the points where the curve shifts direction across the axis.
For the equation \(\frac{x^2}{16} - \frac{y^2}{4} = 1\), substituting \(y = 0\) gives \(\frac{x^2}{16} = 1\). Solving this, we find \(x^2 = 16\), which gives the solutions \(x = \pm 4\). These calculations show that the hyperbola crosses the x-axis at \((4, 0)\) and \((-4, 0)\). These intercepts are critical for plotting the hyperbola accurately, as they define the points where the curve shifts direction across the axis.
Standard Form of Conics
In mathematics, conic sections include circles, ellipses, parabolas, and hyperbolas. Each has a distinct standard form. Recognizing these forms helps in identifying the conic type and its properties.
For hyperbolas, the standard form is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for horizontal hyperbolas, and \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for vertical ones. These forms highlight the subtraction operation, which distinguishes hyperbolas from circles and ellipses.
For hyperbolas, the standard form is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for horizontal hyperbolas, and \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) for vertical ones. These forms highlight the subtraction operation, which distinguishes hyperbolas from circles and ellipses.
- Circle: \(x^2 + y^2 = r^2\)
- Ellipse: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
- Parabola: \(y = ax^2 + bx + c\)
Other exercises in this chapter
Problem 39
Identify whether equation, when graphed, will be a parabola, circle, ellipse, or hyperbola. Sketch the graph of equation. If a parabola, label the vertex. If a
View solution Problem 39
For Exercises 37 through \(42,\) first divide the equation through by the coefficient of \(\left.x^{2} \text { (or } y^{2}\right)\) $$6(x-4)^{2}+6(y-1)^{2}=24$$
View solution Problem 40
For Exercises 37 through \(42,\) first divide the equation through by the coefficient of \(\left.x^{2} \text { (or } y^{2}\right)\) $$7(x-1)^{2}+7(y-3)^{2}=63$$
View solution Problem 41
Find each function value if \(f(x)=3 x^{2}-2 .\) See Section 3.2 $$ f(-1) $$
View solution