Problem 42
Question
Use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes \((y-2)^{2}-(x+3)^{2}=5\)
Step-by-Step Solution
Verified Answer
The center of the hyperbola is at (-3, 2). The vertices are located at \((-3, 2 - \sqrt{5})\) and \((-3, 2 + \sqrt{5})\). The foci are at \((-3, 2 - \sqrt{10})\) and \((-3, 2 + \sqrt{10})\). The equations of the asymptotes are \(y = x + 5\) and \(y = -x -1\).
1Step 1: Identify the center of the hyperbola
The general form of a hyperbola equation is \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\) for a hyperbola opening upward and downward, or \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\) for a hyperbola opening right and left, where (h, k) are the coordinates of the center. From the given equation \((y-2)^2 - (x+3)^2 = 5\), it can be seen that the center of the hyperbola is at (-3, 2).
2Step 2: Determine the Vertices
The distance from the center to each vertex is 'a', which corresponds to the square root of the value on the same side of the equation as the positive term. In this case, \(a = \sqrt{5}\), so the vertices are \((-3, 2 - \sqrt{5})\) and \((-3, 2 + \sqrt{5})\) .
3Step 3: Find the Foci
The distance from the center to each focus is denoted 'c'. The relationship between a, b, and c in a hyperbola is \(c = \sqrt{a^2 + b^2}\). In this case, since \(b^2 = a^2 = 5\), it can be determined that the foci are located at \((-3, 2 - \sqrt{10})\) and \((-3, 2 + \sqrt{10})\) .
4Step 4: Determine the Asymptotes
The asymptotes of a hyperbola are given by the equation \(y = k \pm (a/b)(x - h)\) . In this case, since a = b, the equation of the asymptotes is \(y = 2 \pm (x + 3)\), or \(y = x + 5\) and \(y = -x -1\). These lines define the 'path' that the hyperbola approaches.
5Step 5: Graph the Hyperbola
Plot the center, vertices, and foci found in steps 1-3, and draw the asymptotes found in the step 4. Then, sketch the hyperbola opening toward the positive and negative y-directions, approaching but never crossing the asymptote lines.
Key Concepts
Center of the HyperbolaVertices of the HyperbolaFoci of the HyperbolaAsymptotes of the Hyperbola
Center of the Hyperbola
When graphing a hyperbola, identifying the center is your starting point. The center of a hyperbola in the equation \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\) is represented by the point \((h, k)\). In this case, from \((y-2)^2 - (x+3)^2 = 5\), we rewrite it in standard form by dividing everything by 5, which gives us: \((y-2)^2/5 - (x+3)^2/5 = 1\). Here the center is \((-3, 2)\). This point is the midpoint of the hyperbola, around which all other elements such as vertices and foci are symmetrically distributed.
Vertices of the Hyperbola
Vertices are crucial in determining the shape and direction of a hyperbola. For the equation \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\), the hyperbola opens upward and downward. The distance from the center to each vertex is denoted by 'a'. In our example, we find \(a = \sqrt{5}\). This means the vertices are located at \((-3, 2 - \sqrt{5})\) and \((-3, 2 + \sqrt{5})\). These points lie on the vertical line through the center and define the narrowest points of the hyperbola’s branches.
Foci of the Hyperbola
Foci are points located just outside the vertices, adding depth to the hyperbola's shape. The distance from the center to each focus is 'c', calculated using \(c = \sqrt{a^2 + b^2}\). In this example, \(a^2 = b^2 = 5\), so \(c = \sqrt{5 + 5} = \sqrt{10}\). This places the foci at \((-3, 2 - \sqrt{10})\) and \((-3, 2 + \sqrt{10})\). These points are important for sketching the hyperbola as they define its orientation and extent along the axis in the direction it opens.
Asymptotes of the Hyperbola
Asymptotes are lines that the hyperbola approaches but never touches, providing a guide for its shape. For a hyperbola given by \((y-k)^2/a^2 - (x-h)^2/b^2 = 1\), the asymptotes have equations \(y = k \pm (a/b)(x - h)\). With \(a = b\) in our case, the asymptotes simplify to \(y = 2 \pm (x + 3)\), resulting in lines \(y = x + 5\) and \(y = -x - 1\). These diagonal lines create a framework within which the branches of the hyperbola fit, ensuring its curved trajectory.
Other exercises in this chapter
Problem 41
Graph each ellipse and give the location of its foci. $$\frac{(x-4)^{2}}{9}+\frac{(y+2)^{2}}{25}=1$$
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Find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$(y-1)^{2}=-8 x$$
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Graph each ellipse and give the location of its foci. $$\frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1$$
View solution Problem 43
Convert each equation to standard form by completing the square on \(x\) or \(y .\) Then find the vertex, focus, and directrix of the parabola. Finally, graph t
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