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Chapter 6: Problem 59

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$100 x^{2}+100 y^{2}-100 x+400 y+409=0$$

Short Answer

Expert verified
The graph of the given equation is a circle with center at \((0.5, -2)\) and radius approximately 0.953

Step by step solution

01

Grouping the terms

Rewrite the given equation by grouping the terms related to x and y separately, resulting in: \(100(x^{2} - x) + 100(y^{2} + 4y) = -409\)
02

Completing the square

Complete the square for both x and y terms. For the x terms, the square is completed by adding \((\frac{1}{2} * 1)^{2} = 0.25\) which gives: \(100(x - 0.5)^{2}\). For the y terms, square is completed by adding \((\frac{1}{2} * 4)^{2} = 4\), which results in \(100(y + 2)^{2}\). Therefore, the equation becomes: \(100(x - 0.5)^{2} + 100(y + 2)^{2} = -409 + 100 + 400\) that simplifies to \(100(x - 0.5)^{2} + 100(y + 2)^{2} = 91 \)
03

Identifying the conic section

Now the equation is presented in the standard form of a conic section, which is \((x - h)^{2}/a^{2} + (y - k)^{2}/b^{2} = 1\). The coefficients of \(x^{2}\) and \(y^{2}\) are the same, which suggests that the graph of this equation is a circle with center \((h, k)\) and radius \(r = \sqrt{91/100}\).

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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. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Parabola } & (10, \pi / 2) \\ \end{array} $$

\(r^{2}=36 \sin 2 \theta\)

True or False? In Exercises 59-61, determine whether the statement is true or false. Justify your answer. The conic represented by the following equation is an ellipse. $$ r^{2}=\frac{16}{9-4 \cos \left(\theta+\frac{\pi}{4}\right)} $$

In Exercises 73-78, solve the trigonometric equation. $$ 9 \csc ^{2} x-10=2 $$

Sketch the graph of each equation. (a) \(r=1-\sin \theta\) (b) \(r=1-\sin \left(\theta-\frac{\pi}{4}\right)\)
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