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Chapter 10: Q. 21 (page 683)

In Problems 13鈥�24, analyze each equation and graph it

r(3-2sin胃)=6

Short Answer

Expert verified

The equation represents an ellipse.

Step by step solution

01

Step 1. Given Information 

We are given an equation r(3-2sin胃)=6

We need to find the graph of the equation.

02

Step 2. Analyze the equation 

Compare the given equation with r=ep1-esin胃

The given equation becomes,

r=63-2sin胃r=63-2sin胃路1313r=21-23sin胃OnComparing,e=23<1

Therefore, it is an ellipse.

03

Step 3. Plotting points 

Now, we will calculate the points from the given equation,

胃=0:r=63-2sin(0)=63-0=63=2胃=蟺2:r=63-2sin蟺2=63-2=61=6胃=蟺:r=63-2sin(蟺)=63-0=63=2胃=3蟺2:r=63-2sin3蟺2=63+2=6565,3蟺2

04

Step 4. Required graph 

Plotting the points, we get the graph as,

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Most popular questions from this chapter

True or False

To eliminate the xy-term from the equation x2-2xy+y2-2x+3y+5=0, rotate the axes through an angle 胃, where cot胃=B2-4AC.

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The graph of a parabola is given. Match graph to its equation.

and equations of the parabola areA)y2=4xC)y2=-4xE)(y-1)2=4(x-1)G)(y-1)2=-4(x-1)B)x2=4yD)x2=-4yF)(x+1)2=4(y+1)H)(x+1)2=4(y+1)

To complete the square of x2-3x , add______ .

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Vertex is at4,-2and focus is at6,-2.
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