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Chapter 6: Problem 50

Classify the graph of the equation as a circle, a parabola, an ellipse, or a hyperbola. $$x^{2}+y^{2}-4 x+6 y-3=0$$

Short Answer

Expert verified
The graph of the given equation is a Circle with center at (2,-3) and radius 4.

Step by step solution

01

Identify and arrange the given equation

The given equation is \( x^{2} + y^{2} - 4x + 6y - 3 = 0 \). Try to arrange it in the form of a circle, \( (x-h)^2 + (y-k)^2 = r^2 \). To do this, group the \(x\) terms together and the \(y\) terms together. This results in the equation, \( (x^2 - 4x) + (y^2 + 6y) = 3 \).
02

Complete the square for x and y

To complete the squares, add and subtract \( \frac{(-4)^2}{4} = 4 \) inside the \(x\) group and add and subtract \( \frac{6^2}{4} = 9 \) inside the \(y\) group. Make sure to balance the equation by adding the same value to the other side that is added to the left side. The equation becomes \( (x^2 - 4x + 4) - 4 + (y^2 + 6y + 9) - 9 = 3+4+9 \) which simplifies into \( (x-2)^2 + (y+3)^2 = 16 \).
03

Determine the type of conic section

As the equation is in the form \( (x-h)^2 + (y-k)^2 = r^2 \), the given equation represents a circle.

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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. $$ \begin{array}{lll} {\text { Conic }} & \text { Eccentricity } & \text { Directrix } \\ \text { Hyperbola } & e=2 & x=1 \\ \end{array} $$

In Exercises 33-48, find a polar equation of the conic with its focus at the pole. $$ \begin{array}{ll} {\text { Conic }} & \text { Vertex or Vertices } \\ \text { Hyperbola } & (4, \pi / 2),(1, \pi / 2) \end{array} $$

\(r=2 \csc \theta+5\)

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

In Exercises 29-32, use a graphing utility to graph the rotated conic. $$ r=\frac{5}{-1+2 \cos (\theta+2 \pi / 3)} $$
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