Polar Equations for Conics

Identify the conic that the polar equation represents. Also, give the position of the directrix. 9 1) r = 1 - 3 cos θ A) hyperbola, directrix perpendicular to the polar axis 3 left of the pole B) ellipse, directrix perpendicular to the polar axis 3 left of the pole C) ellipse, directrix perpendicular to the polar axis 3 right of the pole D) hyperbola, directrix perpendicular to the polar axis 3 right of the pole

2 2) r = 2 + 2 sin θ A) hyperbola, directrix parallel to the polar axis 1 above the pole B) parabola, directrix parallel to the polar axis 1 above the pole C) hyperbola, directrix perpendicular to the polar axis 1 right of the pole D) parabola, directrix perpendicular to the polar axis 1 right of the pole

2 3) r = 4 - 2 sin θ A) ellipse, directrix perpendicular to the polar axis 1 right of the pole B) ellipse, directrix parallel to the polar axis 1 above the pole C) ellipse, directrix parallel to the polar axis 1 below the pole D) ellipse, directrix perpendicular to the polar axis 1 left of the pole

6 4) r = 3 - 4 cos θ 3 A) ellipse, directrix is perpendicular to the polar axis at a distance units to the right of the pole 2 B) hyperbola, directrix is perpendicular to the polar axis at a distance 3 units to the right of the pole 3 C) hyperbola, directrix is perpendicular to the polar axis at a distance units to the left of the pole 2 D) ellipse, directrix is perpendicular to the polar axis at a distance 3 units to the left of the pole

Discuss the equation and graph it. 6 5) r = 3 - 3 cos θ

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5 PreCalculus

9 6) r = 3 - sin θ

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5

3 7) r = 2 + 4 sin θ

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5

Find a polar equation for the conic. A focus is at the pole. 8) e = 1; directrix is parallel to the polar axis 2 above the pole 2 2 2 2 A) r = B) r = C) r = D) r = 1 + sin θ 1 + cos θ 1 - cos θ 1 - sin θ

1 9) e = ; directrix is perpendicular to the polar axis 2 to the left of the pole 2 8 4 4 8 A) r = B) r = C) r = D) r = 4 - 2 cos θ 4 + 2 cos θ 4 - 2 cos θ 4 + 2 cos θ

Convert the polar equation to a rectangular equation. 4 10) r = 2 - 2 cos θ A) x2 = -4y + 4B)x2 = 4y + 4C)y2 = -4x + 4D)y2 = 4x + 4

6 11) r = 2 - 2 sinθ A) x2 = 6y + 9B)x2 = -6y + 9C)y2 = 6x + 9D)y2 = -6x + 9

Calin M. Agut - 2012 PreCalculus

6 12) r = 2 + cos θ A) 5x2 + 4y2 - 12x - 36 = 0B)4x2 + 3y2 + 12y - 36 = 0 C) 4x2 + 4y2 + 12x - 36 = 0D)3x2 + 4y2 + 12x - 36 = 0

8 13) r = 2 + sin θ A) 3x2 + 4y2 + 16x - 64 = 0B)4x2 + 3y2 + 16y - 64 = 0 C) 5x2 + 4y2 - 16x - 64 = 0D)4x2 + 4y2 + 16x - 64 = 0

14) r(3 + cos θ)= 6 A) 9x2 + 8y2 + 12y - 36 = 0B)9x2 + 9y2 + 12x - 36 = 0 C) 10x2 + 9y2 - 12x - 36 = 0D)8x2 + 9y2 + 12x - 36 = 0

6 sec θ 15) r = 3 sec θ + 1 A) 9x2 + 9y2 + 12x - 36 = 0B)8x2 + 9y2 + 12x - 36 = 0 C) 9x2 + 8y2 + 12y - 36 = 0D)10x2 + 9y2 - 12x - 36 = 0

Calin M. Agut - 2012 Answer Key Testname: 15_POLAR_EQ_CONICS

1) A 2) B 3) C 4) C 5) parabola; directrix perpendicular to the polar axis 2 units to left of pole focus (0, 0), vertex 1, π

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5

6) ellipse; directrix parallel to the polar axis 9 units below pole 9 π center , 8 2 9 π 9 3π vertices , , , 2 2 4 2

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5

Calin M. Agut - 2012 Answer Key Testname: 15_POLAR_EQ_CONICS

7) hyperbola; directrix parallel to 3 the polar axis unit above the pole 4 1 π 3 3π vertices , , - , 2 2 2 2

5 4 3 2 1

-5 -4 -3 -2 -1 1 2 3 4 5 r -1 -2 -3 -4 -5 8) A 9) C 10) D 11) A 12) D 13) B 14) D 15) B

Calin M. Agut - 2012