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Find the center, foci, vertices, asymptotes, and radius, as appropriate, of the conic sections in Exercises \(57-68 .\) $$ x^{2}-y^{2}-2 x+4 y=4 $$

Short Answer

Expert verified
Center: (1, 2); Vertices: (4, 2), (-2, 2); Foci: (1±3√2, 2); Asymptotes: y = x + 1, y = -x + 3.

Step by step solution

01

Rewrite the Equation in Standard Form

To find the center, foci, vertices, asymptotes, and radius of this hyperbola, we start by rewriting the given equation x^{2}-y^{2}-2x+4y=4 into its standard form. Group the x-terms and y-terms: \( (x^2 - 2x) - (y^2 - 4y) = 4 \).Complete the square for the x-terms and y-terms:- For \(x\) terms: \(x^2 - 2x\), complete to \((x - 1)^2 - 1\).- For \(y\) terms: \(y^2 - 4y\), complete to \((y - 2)^2 - 4\).Substitute back: \((x - 1)^2 - 1 - ((y - 2)^2 - 4) = 4\), which simplifies to \((x - 1)^2 - (y - 2)^2 = 9\).
02

Identify the Center and Axes

From the standard form of a hyperbola, \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), we recognize the equation as \((x-1)^2/9 - (y-2)^2/9 = 1\). Here, \( h = 1 \) and \( k = 2 \), so the center of the hyperbola is \((1, 2)\). The principal axes are aligned with the x and y-axes.
03

Determine the Vertices

The vertices of a hyperbola lie along the transverse axis, directly from the center to both positive and negative direction. Since \( a^2 = 9 \), \( a = 3 \). The vertices are therefore located 3 units away from the center along the x-direction, giving points \((h+a, k) = (4, 2)\) and \((h-a, k) = (-2, 2)\).
04

Calculate the Foci

For a hyperbola, the foci are located using \(c^2 = a^2 + b^2\). Since \(a^2 = 9\) and \(b^2 = 9\), \(c^2 = 18\) yielding \(c = \sqrt{18} = 3\sqrt{2}\).The foci are found at \((h+c, k) = (1+3\sqrt{2}, 2)\) and \((h-c, k) = (1-3\sqrt{2}, 2)\).
05

Find the Asymptotes

The asymptotes of a hyperbola are straight lines that pass through the center and align with the asymptotic direction defined by the slopes \(\pm b/a\).For this hyperbola, the asymptotes are: - \(y - k = \pm (b/a)(x - h)\)- Substituting \(b=3\), \(a=3\), we get \(y - 2 = \pm(x - 1)\), resulting in equations: \(y = x + 1\) and \(y = -x + 3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Hyperbolas
Hyperbolas are fascinating conic sections distinguished by their characteristic shape. They consist of two separate curves and are defined by subtracting the square of one variable from another. The general equation of a hyperbola is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] where \((h, k)\) is the center of the hyperbola.
Unlike circles and ellipses, where we see a cohesive curve, hyperbolas are split into two separate branches. Each branch approaches two lines called "asymptotes," but never intersects them. Depending on the positions of \(x\) and \(y\) in the equation, a hyperbola can open either horizontally or vertically. In our solved equation, it opens horizontally because \(x\) is positive.
The beauty and complexity of hyperbolas make them a rich topic in geometry, with applications in navigation, astronomy, and even in architecture.
Equation of Asymptotes
For hyperbolas, asymptotes are crucial features. They are the invisible lines that the hyperbola approaches but never touches. The equations of these lines provide deep insight into the hyperbola's structure.
The formula for calculating the asymptotes' equations from the standard hyperbola form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) is given by:
\[ y - k = \pm \frac{b}{a}(x - h) \]
In the given exercise, where both \(a\) and \(b\) are 3, the slope \(\frac{b}{a}\) equals 1. This results in the equations of the asymptotes being \(y = x + 1\) and \(y = -x + 3\).
These straight lines establish the boundaries within which the hyperbola's branches stretch out.
Vertices of a Hyperbola
The vertices of a hyperbola are significant points where the curves are closest to each other. These points are on the transverse axis, the line that passes through both the hyperbola's center and its vertices.
To find these vertices, you use the value of \(a\) from the hyperbola's standard form equation.
For our problem, where \(a^2=9\), the value of \(a\) is 3. The vertices are therefore located a distance of 3 units along the x-axis from the center \((1,2)\), resulting in vertices at \((4,2)\) and \((-2,2)\).
These vertices define the extent of the transverse axis and are integral features of the hyperbola's structure. Understanding the location of vertices helps grasp the complete geometry of the hyperbola.
Foci of a Hyperbola
The foci are another set of defining points of a hyperbola. They are located along the transverse axis, outside the vertices. The importance of foci lies in their role in the geometric definition: for any point on the hyperbola, the absolute difference of its distances to the foci is constant.
To find the foci, we calculate \(c\) using the formula \(c^2 = a^2 + b^2\). With our values of \(a^2 = 9\) and \(b^2 = 9\), we have \(c^2 = 18\), giving \(c = 3\sqrt{2}\).
Thus, the foci are positioned at \((1 + 3\sqrt{2}, 2)\) and \((1 - 3\sqrt{2}, 2)\). These points emphasize the open curves of the hyperbola and play a key role in various scientific and engineering calculations.

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Most popular questions from this chapter

Graph the lines and conic sections in Exercises \(47-56\) $$ r=4 \sin \theta $$

The "triangular" region in the first quadrant bounded by the \(x\) -axis, the line \(x=4,\) and the hyperbola \(9 x^{2}-4 y^{2}=36\) is revolved about the \(x\) -axis to generate a solid. Find the volume of the solid.

Use a CAS to plot the polar equation $$ r=\frac{k e}{1+e \cos \theta} $$ for various values of \(k\) and \(e,-\pi \leq \theta \leq \pi .\) Answer the following questions. a. Take \(k=-2 .\) Describe what happens to the plots as you take \(e\) to be \(3 / 4,1,\) and \(5 / 4 .\) Repeat for \(k=2 .\) b. Take \(k=-1 .\) Describe what happens to the plots as you take \(e\) to be \(7 / 6,5 / 4,4 / 3,3 / 2,2,3,5,10,\) and \(20 .\) Repeat for \(e=1 / 2,1 / 3,1 / 4,1 / 10,\) and \(1 / 20 .\) c. Now keep \(e>0\) fixed and describe what happens as you take \(k\) to be \(-1,-2,-3,-4,\) and \(-5 .\) Be sure to look at graphs for parabolas, ellipses, and hyperbolas.

The region bounded on the left by the \(y\) -axis, on the right by the hyperbola \(x^{2}-y^{2}=1,\) and above and below by the lines \(y=\pm 3\) is revolved about the \(y\) -axis to generate a solid. Find the volume of the solid.

The hyperbola \(\left(x^{2} / 16\right)-\left(y^{2} / 9\right)=1\) is shifted 2 units to the right to generate the hyperbola $$\frac{(x-2)^{2}}{16}-\frac{y^{2}}{9}=1$$ a. Find the center, foci, vertices, and asymptotes of the new hyperbola. b. Plot the new center, foci, vertices, and asymptotes, and sketch in the hyperbola.

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