Problem 14
Question
Use vertices and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. \(\frac{x^{2}}{16}-\frac{y^{2}}{25}-1\)
Step-by-Step Solution
Verified Answer
The foci of the hyperbola are at (±√41,0). The equations of the asymptotes are y = ±(5/4)x. The hyperbola's vertices are at (±4, 0).
1Step 1: Role of Constants a and b
In the equation \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\), a = 4 and b = 5 are the semi-major and semi-minor axis lengths respectively of the hyperbola.
2Step 2: Find the value of c
c is the distance from the center to a focus and is found using the equation \(c = \sqrt{a^{2}+b^{2}}\). Substituting a = 4 and b = 5, we get \(c = \sqrt{4^{2}+5^{2}}= \sqrt{41}\). Our foci will be located at (±c , 0), so Foci = (± √41, 0).
3Step 3: Formulation of Asymptote Equations
Since the x term has the positive coefficient, the equations of the asymptotes for a hyperbola in this form will be y = ±(b/a)x which gives us the equations y = ±(5/4)x.
4Step 4: Graphing the Hyperbola along with its Asymptotes
The vertices are at points (±a,0) or (±4,0). Use this and the foci to plot the hyperbola. The asymptotes will be the lines passing through the origin making an angle of \(tan^{-1}(\frac{b}{a})\) or \(tan^{-1}(\frac{5}{4})\) with the x-axis. The graph of the hyperbola will approach these lines as the variable x or y tends to infinity.
Key Concepts
VerticesAsymptotesFociSemi-Major Axis
Vertices
In a hyperbola, the vertices are crucial points that define its shape and position. The vertices lie on the transverse axis, which is the axis intersected by the hyperbola itself.
- The standard form of the horizontal hyperbola equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
- For this equation, the vertices are at \((\pm a, 0)\).
- The vertices are located at \((4, 0)\) and \((-4, 0)\).
Asymptotes
Asymptotes are the lines that a hyperbola approaches but never actually touches. They serve as guiding boundaries that help in understanding the behavior and spread of the hyperbola.
- In the standard form equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations of the asymptotes are \( y = \pm \frac{b}{a}x \).
- From our equation, this gives \( y = \pm \frac{5}{4}x \).
Foci
The foci are specific points that are located along the transverse axis, further from the center than the vertices. The distance to the foci is determined by a relationship involving the semi-major and semi-minor axes lengths.
- The formula for finding the distance to the foci from the center is \( c = \sqrt{a^2 + b^2} \).
- For our hyperbola, substituting \( a = 4 \) and \( b = 5 \) results in \( c = \sqrt{41} \).
Semi-Major Axis
In the context of hyperbolas, the semi-major axis is one of the determining factors of its size and is usually the longer axis of the conic. However, unlike an ellipse, in a hyperbola the term 'semi-major axis' can be misleading, as the hyperbola does not enclose the axis.
- In this particular hyperbola, \( a = 4 \) and \( b = 5 \), designating \( b \) as the semi-major axis since \( b > a \).
- It describes the distance from the center to a vertex along the imaginary axis of the hyperbola.
Other exercises in this chapter
Problem 13
Graph each ellipse and locate the foci. $$ 25 x^{2}+4 y^{2}=100 $$
View solution Problem 14
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. $$x^{2}-6 y=0$$
View solution Problem 14
Graph each ellipse and locate the foci. $$ 9 x^{2}+4 y^{2}=36 $$
View solution Problem 15
Find the focus and directrix of the parabola with the given equation. Then graph the parabola. $$8 x^{2}+4 y=0$$
View solution