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Chapter 6: Problem 60

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$4 x^{2}-y^{2}+4 x+2 y-1=0$$

Short Answer

Expert verified
The graph of the equation \(4x^{2} - y^{2} + 4x + 2y - 1 = 0\) is a hyperbola.

Step by step solution

01

Analyze the Equation

Analyzing the equation \(4x^{2}-y^{2}+4x+2y-1=0\), it can be observed that both \(x^{2}\) and \(y^{2}\) terms are present. Moreover, these terms have different coefficients (4 and -1 respectively) and different signs. This hints towards the graph being a hyperbola.
02

Rearrange the Equation

Next, reorder the terms and group the like ones together to try to bring the equation into standard form. The equation can be written as \(4(x^{2}+x) - (y^{2}-2y) = 1\). This requires further rearrangement to fit the standard forms of conic sections equations.
03

Complete the Square

Completing the square for the \(x\) and \(y\) terms will help to bring the equation into a standard form. For the \(x\)-terms, recall that \((a+b)^2 = a^2 + 2ab + b^2\). Setting \(4(x^{2} + x) = 4(x + 0.5)^2 - 1\), we achieve this. Similarly, for the \(y\)-terms, the same method yields \((y^{2} - 2y) = (y - 1)^2 - 1\). Substituting these back into the equation gives \(4(x + 0.5)^2 - (1 - y)^2 = 2\)
04

Final format and conclusion

Finally, divide the entire equation by 2, which gets it into the standard form of a hyperbola, which is \(\frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1\). The final equation is \(2(x + 0.5)^2 - (1 - y)^2 = 1\). As suspected from the analysis, the graph is a hyperbola since the \(x^{2}\) and \(y^{2}\) terms are subtracted and their coefficients are unequal.

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Most popular questions from this chapter

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Parabola } & (10, \pi / 2) \\ \end{array} $$

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Ellipse } & (2, \pi / 2),(4,3 \pi / 2) \\ \end{array} $$

\(r=4+3 \cos \theta\)

Foci: \((3,2),(3,-4)\); major axis of length 8

Consider the equation \(r=3 \sin k \theta\). (a) Use a graphing utility to graph the equation for \(k=1.5\). Find the interval for \(\theta\) over which the graph is traced only once. (b) Use a graphing utility to graph the equation for \(k=2.5\). Find the interval for \(\theta\) over which the graph is traced only once. (c) Is it possible to find an interval for \(\theta\) over which the graph is traced only once for any rational number \(k\) ? Explain.
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