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For the following exercises, convert the polar equation of a conic section to a rectangular equation.

r = 4 1 + 3   sin   θ

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r = 2 5 3   sin   θ

25 x 2 + 16 y 2 12 y 4 = 0

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r = 8 3 2   cos   θ

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r = 3 2 + 5   cos   θ

21 x 2 4 y 2 30 x + 9 = 0

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r = 4 2 + 2   sin   θ

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r = 3 8 8   cos   θ

64 y 2 = 48 x + 9

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r = 2 6 + 7   cos   θ

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r = 5 5 11   sin   θ

96 y 2 25 x 2 + 110 y + 25 = 0

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r ( 5 + 2   cos   θ ) = 6

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r ( 2 cos   θ ) = 1

3 x 2 + 4 y 2 2 x 1 = 0

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r ( 2.5 2.5   sin   θ ) = 5

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r = 6 sec   θ 2 + 3   sec   θ

5 x 2 + 9 y 2 24 x 36 = 0

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r = 6 csc   θ 3 + 2   csc   θ

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For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci.

r = 2 3 + 3   sin   θ

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r = 10 5 4   sin   θ

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r = 3 1 + 2   cos   θ

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r = 8 4 5   cos   θ

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r = 3 4 4   cos   θ

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r = 6 3 + 2   sin   θ

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r ( 3 4 sin   θ ) = 9

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r ( 3 2 sin   θ ) = 6

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r ( 6 4 cos   θ ) = 5

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For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix.

Directrix: x = 4 ; e = 1 5

r = 4 5 + cos θ

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Directrix: x = 4 ; e = 5

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Directrix: y = 2 ; e = 2

r = 4 1 + 2 sin θ

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Directrix: y = 2 ; e = 1 2

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Directrix: x = 1 ; e = 1

r = 1 1 + cos θ

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Directrix: x = 1 ; e = 1

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Directrix: x = 1 4 ; e = 7 2

r = 7 8 28 cos θ

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Directrix: y = 2 5 ; e = 7 2

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Directrix: y = 4 ; e = 3 2

r = 12 2 + 3 sin θ

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Directrix: x = −2 ; e = 8 3

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Directrix: x = −5 ; e = 3 4

r = 15 4 3 cos θ

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Directrix: y = 2 ; e = 2.5

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Directrix: x = −3 ; e = 1 3

r = 3 3 3 cos θ

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Extensions

Recall from Rotation of Axes that equations of conics with an x y term have rotated graphs. For the following exercises, express each equation in polar form with r as a function of θ .

x 2 + x y + y 2 = 4

r = ± 2 1 + sin θ cos θ

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2 x 2 + 4 x y + 2 y 2 = 9

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16 x 2 + 24 x y + 9 y 2 = 4

r = ± 2 4 cos θ + 3 sin θ

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Chapter review exercises

The Ellipse

For the following exercises, write the equation of the ellipse in standard form. Then identify the center, vertices, and foci.

x 2 25 + y 2 64 = 1

x 2 5 2 + y 2 8 2 = 1 ; center: ( 0 , 0 ) ; vertices: ( 5 , 0 ) , ( −5 , 0 ) , ( 0 , 8 ) , ( 0 , 8 ) ; foci: ( 0 , 39 ) , ( 0 , 39 )

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( x 2 ) 2 100 + ( y + 3 ) 2 36 = 1

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9 x 2 + y 2 + 54 x 4 y + 76 = 0

( x + 3 ) 2 1 2 + ( y 2 ) 2 3 2 = 1 ( 3 , 2 ) ; ( 2 , 2 ) , ( 4 , 2 ) , ( 3 , 5 ) , ( 3 , 1 ) ; ( 3 , 2 + 2 2 ) , ( 3 , 2 2 2 )

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9 x 2 + 36 y 2 36 x + 72 y + 36 = 0

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For the following exercises, graph the ellipse, noting center, vertices, and foci.

x 2 36 + y 2 9 = 1

center: ( 0 , 0 ) ; vertices: ( 6 , 0 ) , ( −6 , 0 ) , ( 0 , 3 ) , ( 0 , −3 ) ; foci: ( 3 3 , 0 ) , ( 3 3 , 0 )

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( x 4 ) 2 25 + ( y + 3 ) 2 49 = 1

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4 x 2 + y 2 + 16 x + 4 y 44 = 0

center: ( −2 , −2 ) ; vertices: ( 2 , −2 ) , ( −6 , −2 ) , ( −2 , 6 ) , ( −2 , −10 ) ; foci: ( −2 , −2 + 4 3 , ) , ( −2 , −2 −4 3 )

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2 x 2 + 3 y 2 20 x + 12 y + 38 = 0

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For the following exercises, use the given information to find the equation for the ellipse.

Center at ( 0 , 0 ) , focus at ( 3 , 0 ) , vertex at ( −5 , 0 )

x 2 25 + y 2 16 = 1

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Center at ( 2 , −2 ) , vertex at ( 7 , −2 ) , focus at ( 4 , −2 )

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A whispering gallery is to be constructed such that the foci are located 35 feet from the center. If the length of the gallery is to be 100 feet, what should the height of the ceiling be?

Approximately 35.71 feet

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The Hyperbola

For the following exercises, write the equation of the hyperbola in standard form. Then give the center, vertices, and foci.

( y + 1 ) 2 16 ( x 4 ) 2 36 = 1

( y + 1 ) 2 4 2 ( x 4 ) 2 6 2 = 1 ; center: ( 4 , −1 ) ; vertices: ( 4 , 3 ) , ( 4 , −5 ) ; foci: ( 4 , −1 + 2 13 ) , ( 4 , −1 2 13 )

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9 y 2 4 x 2 + 54 y 16 x + 29 = 0

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3 x 2 y 2 12 x 6 y 9 = 0

( x 2 ) 2 2 2 ( y + 3 ) 2 ( 2 3 ) 2 = 1 ; center: ( 2 , −3 ) ; vertices: ( 4 , −3 ) , ( 0 , −3 ) ; foci: ( 6 , −3 ) , ( −2 , −3 )

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For the following exercises, graph the hyperbola, labeling vertices and foci.

Questions & Answers

the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
a man walks up 200 meters along a straight road whose inclination is 30 degree.How high above the starting level is he?
Lhorren Reply
100 meters
Kuldeep
Find that number sum and product of all the divisors of 360
jancy Reply
answer
Ajith
exponential series
Naveen
what is subgroup
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Prove that: (2cos&+1)(2cos&-1)(2cos2&-1)=2cos4&+1
Macmillan Reply
e power cos hyperbolic (x+iy)
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10y
Michael
tan hyperbolic inverse (x+iy)=alpha +i bita
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prove that cos(π/6-a)*cos(π/3+b)-sin(π/6-a)*sin(π/3+b)=sin(a-b)
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why {2kπ} union {kπ}={kπ}?
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why is {2kπ} union {kπ}={kπ}? when k belong to integer
Huy
if 9 sin theta + 40 cos theta = 41,prove that:41 cos theta = 41
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what is complex numbers
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Solve 2cos x + 3sin x = 0.5
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Practice Key Terms 2

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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