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Magazine Chapter 14: Problem 1
In Exercises 1 through 6 , find the eccentricity, center, foci, and directrices of each of the given ellipses and draw a sketch of the graph. $$ 6 x^{2}+9 y^{2}-24 x-54 y+51=0 $$
Short Answer
Expert verified
Eccentricity: \(\frac{\sqrt{6}}{2}\), Center: (-3, -1), Foci: \((-3 \pm 2\sqrt{6}, -1)\), Directrices: \(x = -3 \pm \frac{4\sqrt{6}}{3}\), Asymptotes: \( y + 1 = \pm \frac{\sqrt{2}}{2}(x + 3) \).
Step by step solution
01
Rewrite the Equation
Rewrite the given equation in the standard form of a hyperbola. Given:\[ 9x^2 - 18y^2 + 54x - 36y + 79 = 0 \] First, combine and complete the squares:\[ 9(x^2 + 6x) - 18(y^2 + 2y) + 79 = 0 \]Complete the squares:\[ 9(x^2 + 6x + 9 - 9) - 18(y^2 + 2y + 1 - 1) + 79 = 0 \]Simplify:\[ 9((x + 3)^2 - 9) - 18((y + 1)^2 - 1) + 79 = 0 \]Distribute and combine constants:\[ 9(x + 3)^2 - 81 - 18(y + 1)^2 + 18 + 79 = 0 \]\[ 9(x + 3)^2 - 18(y + 1)^2 + 16 = 0 \]Move constants to the right side:\[ 9(x + 3)^2 - 18(y + 1)^2 = -16 \] Divide by -16 to normalize the equation:\[ \frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{8} = 1 \]
02
Identify the Center
The equation is now in the standard hyperbola form:\[ \frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{8} = 1 \] The center of the hyperbola is at:\[ (h, k) = (-3, -1) \]
03
Identify the Axes and Eccentricity
Identify the values for a, b, and c. a is the distance along the x-axis: \[ a^2 = 16 \implies a = 4 \]b is the distance along the y-axis: \[ b^2 = 8 \implies b = 2 \sqrt{2} \] The eccentricity e is determined by c, where:\[ c^2 = a^2 + b^2 \] Then solve for c:\[ c^2 = 16 + 8 = 24 \implies c = \sqrt{24} = 2\sqrt{6} \] Finally, the eccentricity e is:\[ e = \frac{c}{a} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2} \]
04
Find the Foci
The foci are located c units from the center along the transverse axis. So, the foci are at points:\[ (-3 + 2\sqrt{6}, -1) \] and\[ (-3 - 2\sqrt{6}, -1) \]
05
Directrices
The directrices are perpendicular to the transverse axis and located at a specific distance from the center:\[ x = -3 \pm \frac{a^2}{c} \implies x = -3 \pm \frac{16}{2\sqrt{6}} = -3 \pm \frac{8\sqrt{6}}{6} = -3 \pm \frac{4\sqrt{6}}{3} \]
06
Equation of the Asymptotes
The asymptotes for a hyperbola are given by the equation:\[ y - k = \pm \frac{b}{a}(x - h) \]Substitute in the values for h, k, a, and b:\[ y + 1 = \pm \frac{2\sqrt{2}}{4}(x + 3) \implies y + 1 = \pm \frac{\sqrt{2}}{2}(x + 3) \]Two equations for the asymptotes result:\[ y + 1 = \frac{\sqrt{2}}{2}(x + 3) \] and\[ y + 1 = -\frac{\sqrt{2}}{2}(x + 3) \]
07
Sketch the Graph
To draw the hyperbola, plot the center at (-3, -1). Draw the asymptotes using the equations found in Step 6. Plot the vertices 4 units to the left and right of the center along the x-axis. Draw branches of the hyperbola opening left and right, approaching the asymptotes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Eccentricity of Hyperbola
The eccentricity, or *e*, of a hyperbola is a measure of how 'stretched' the hyperbola is. It's computed from the semi-major axis (*a*) and the semi-minor axis (*b*). For hyperbolas, it is given by the formula: \[ e = \frac{c}{a} \] where *c* is the distance from the center to each focus, found using c虏 = a虏 + b虏. The given equation is: 9x虏 - 18y虏 + 54x - 36y + 79 = 0. You first rewrite it in the standard form to extract values for *a* and *b*. Once rearranged, you find: a虏 = 16 \ a = 4 \ b虏 = 8 \ b = 2\sqrt{2} \ c虏 = a虏 + b虏 = 16 + 8 = 24 \ c = 2\sqrt{6} \ Finally, to find *e*: e = \frac{c}{a} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2} =~ 1.22. A higher eccentricity means a more elongated shape.
Foci of Hyperbola
The foci of the hyperbola are points located at a distance *c* from the center along the transverse axis. The hyperbola in standard form is: \[ \frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{8} = 1 \] Here, the center is (h, k) = (-3, -1) and c = 2\sqrt{6}. The foci are thus at: \[ ( h \pm c, k) \] Substituting the values: \[ (-3 + 2\sqrt{6}, -1) \] \[ (-3 - 2\sqrt{6}, -1) \] These points help in sketching and understanding how the branches of the hyperbola stretch.
Asymptotes of Hyperbola
The asymptotes of a hyperbola are lines that the hyperbola approaches but never meets. They guide the shape of the hyperbola. For the hyperbola \[ \frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{8} = 1 \], the asymptotes are given by: \[ y - k = \pm \frac{b}{a}(x - h) \] Plug in the values: h = -3, k = -1, a = 4, b = 2\sqrt{2}. Then, you get: \[ y + 1 = \pm \frac{2\sqrt{2}}{4}(x + 3) \] Which simplifies to: \[ y + 1 = \pm \frac{\sqrt{2}}{2}(x + 3) \] This results in two equations: \[ y + 1 = \frac{\sqrt{2}}{2}(x + 3) \] and \[ y + 1 = -\frac{\sqrt{2}}{2}(x + 3) \] These lines pass through the center and give the direction in which the hyperbola opens.
Directrices of Hyperbola
The directrices of a hyperbola are lines that are related to the distance between any point on the hyperbola and its foci. For this hyperbola \[ \frac{(x + 3)^2}{16} - \frac{(y + 1)^2}{8} = 1 \], the directrices are vertical lines perpendicular to the transverse axis, located at a distance of \frac{a^2}{c} from the center: \[ \text{Directrix: } x = h \pm \frac{a^2}{c} \] Substituting for h = -3, a虏 = 16, and c = 2\sqrt{6}, we have: \[ x = -3 \pm \frac{16}{2\sqrt{6}} = -3 \pm \frac{8\sqrt{6}}{6} = -3 \pm \frac{4\sqrt{6}}{3} \] These directrices are key in determining how 'open' the hyperbola is and how it relates to the foci.
