Question

PLEASE HELP FAST!!!! 20 POINTS!!!!

1.) Which of the following polar equations is equivalent to the parametric equations below?
x=t^2
y=2t
A.) r=4cot(theta)csc(theta)
B.) r=4tan(theta)sec(theta)
C.) r=tan(theta)sec(theta)/4
D.) r=16cot(theta)csc(theta)

2.) Which polar equation is equivalent to the parametric equations below? x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)
A.) r=cos(theta)+1
B.) r=sin(theta)+1
C.) r=sin^2(theta)+1
D.) r=1-cos(theta)

Answer 1

Part (1)
x = t² , y = 2t
we should know that ⇒ x = r cos θ , y = r sin θ
By equating the values of x ⇒ ∴ t² = r cos θ →(1)
By equating the values of y ⇒ ∴ 2t = r sin θ ⇒ ∴ 4t² = r² sin² θ → (2)


dividing equation (2) over (1)
4t²/t² = (r² sin² θ)/(r cos θ)
∴ 4 = r (sin² θ/cos θ)
∴ r = 4 cos θ / sin² θ = 4 ( cos θ / sin θ ) ( 1 / sin θ)
∴ r = 4 (cot θ) (csc θ)

The correct answer is option A.) r=4cot(theta) csc(theta)
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Part (2)
x = sin(θ) cos(θ) + cos(θ) , y = sin²(θ) + sin(θ)
we should know that ⇒ x = r cos θ , y = r sin θ

By equating the values of x ⇒
∴ r cos θ = sin(θ) cos(θ) + cos(θ) ⇒⇒ take cos θ as common from right side

∴ r cos θ = cos θ ( 1 + sin θ) ⇒⇒⇒ divide both sides over cos θ

∴ r = 1 + sin θ → (1)


By equating the values of y ⇒
∴ r sin θ = sin²(θ) + sin(θ) ⇒⇒ take sin θ as common from right side

∴ r sin θ = sin θ ( 1 + sin θ) ⇒⇒⇒ divide both sides over sinθ

∴ r = 1 + sin θ → (2)


From (1) and (2)
∴ The correct answer is option B.) r = sin θ + 1