Problem 70
Question
Graph each equation. $$ 4 x^{2}-9 y^{2}=36 $$
Step-by-Step Solution
Verified Answer
The graph of the equation is a hyperbola centered at (0,0), with vertices at (±3,0), foci at (±√13,0), and with asymptotes \(y = ± (2/3)x\).
1Step 1: Set the equation in the standard form
The equation should be written in the standard form for a hyperbola. If a hyperbola opens sideways, the equation is \((x - h)^2 / a^2 - (y - k)^2 / b^2 = 1\). For a hyperbola that opens up and down, the equation is \((y - k)^2 / a^2 - (x - h)^2 / b^2 = 1\). The current equation can be rewritten as \((x^2/3^2) - (y^2/2^2) = 1\). So, from the equation, a = 3, b = 2 and the hyperbola is centered at (h, k) = (0, 0).
2Step 2: Identify vertices and foci
The vertices of the hyperbola are (±a, 0), or (±3, 0). The foci are at (±c, 0) where \(c = \sqrt{a^2 + b^2} = \sqrt{3^2 + 2^2} = \sqrt{13}\). So the coordinates of the foci are \((± \sqrt{13}, 0)\).
3Step 3: Graphing the hyperbola
To graph, start by drawing the center at the origin. Then mark the vertices and foci. Draw the two branches of the hyperbola, making sure they pass through the vertices and approach the asymptotes. The asymptotes are the diagonal lines that pass through the center and establish the path along which the hyperbola approaches as it moves away from the center. For the given equation the asymptotes are \(y = ± (2/3)x\).
Key Concepts
Graphing HyperbolasConic SectionsIdentifying Vertices and FociEquation of HyperbolaAsymptotes of Hyperbola
Graphing Hyperbolas
Hyperbolas are fascinating curves shaped somewhat like two mirrored arches. They have two separate parts, known as branches, that veer off to infinity. When graphing hyperbolas, the first step is to identify the center, vertices, and foci. These elements are crucial in shaping and positioning the hyperbola on a graph.
- Start by plotting the center point.
- Then, plot the vertices along the transverse axis.
- The foci, further away from the center than the vertices, guide the curve of each branch.
Conic Sections
A hyperbola is one of the four conic sections, the others being the circle, ellipse, and parabola. Conic sections result from slicing a cone at different angles. Hyperbolas occur when the cone is sliced with a plane in such a way that it cuts through both halves of the cone.
This dual-branch feature distinguishes hyperbolas from other conic sections:
This dual-branch feature distinguishes hyperbolas from other conic sections:
- Circles and ellipses are closed curves.
- Parabolas are open curves with one branch.
- Hyperbolas have two distinct branches.
Identifying Vertices and Foci
Vertices and foci are fundamental components of hyperbolas. They help define the shape and orientation of the curve. The vertices are the points where each branch is closest together, while the foci lie outside the vertices, along the transverse axis.
For the equation \( (x^2/3^2) - (y^2/2^2) = 1 \):
For the equation \( (x^2/3^2) - (y^2/2^2) = 1 \):
- The vertices are at \((±3, 0)\), indicating the points where the branches cross the x-axis.
- The foci, calculated using \(c = \sqrt{a^2 + b^2} = \sqrt{13}\), are at \((±\sqrt{13}, 0)\).
Equation of Hyperbola
The standard form of a hyperbola's equation reveals its orientation and properties. For hyperbolas centered at the origin, the equations are:
- \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\) for hyperbolas opening sideways.
- \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\) for hyperbolas opening upwards and downwards.
- The transverse axis is horizontal because \(x^2\) comes first in the equation.
- The center is at the origin \((0,0)\).
- By identifying \(a\) and \(b\) from this form, we can plot the vertices and calculate the foci, completing the hyperbola's framework.
Asymptotes of Hyperbola
Asymptotes are lines that the branches of the hyperbola approach but never touch. They play a critical role in graphing because they define the slope and direction of the curve as it extends away from the center. For hyperbolas aligned with the x-axis, the asymptotes form through the origin, following the equation:
- \(y = ±(b/a)x\) for horizontal hyperbolas.
- For our earlier example, this translates to \(y = ±(2/3)x\).
Other exercises in this chapter
Problem 69
Solve each equation. Check your answers. \(\frac{1}{3 x+1}=\frac{1}{x^{2}-3}\)
View solution Problem 70
What is the length of the major axis on the graph of \(\frac{x^{2}}{100}+\frac{y^{2}}{64}=1 ?\) \(\begin{array}{llll}{\text { F. } 12} & {\text { G. } 2 \sqrt{4
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Expand each binomial. $$ (x-y)^{3} $$
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Solve each equation. Check your answers. \(\frac{2}{x+2}=\frac{6}{x^{2}-4}\)
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