Question

Write the following parametric equations as a polar equation.

x equals lower t squared
y = 2t
(1 point)
Responses

r = 4 cot θ csc θ
r = 4 cot θ csc θ

r = 2 cot θ csc θ
r = 2 cot θ csc θ

r = 4 tan θ sec θ
r = 4 tan θ sec θ

r = 2 tan θ sec θ

Answer 1

To convert the given parametric equations \( x = t^2 \) and \( y = 2t \) to a polar equation, we can use the relationships between Cartesian coordinates and polar coordinates:

- \( x = r \cos \theta \)
- \( y = r \sin \theta \)
- \( r = \sqrt{x^2 + y^2} \)

From the equations, we also identify \( t \) in terms of \( y \):
\[ t = \frac{y}{2} \]

Using this value of \( t \), we can express \( x \):
\[ x = t^2 = \left(\frac{y}{2}\right)^2 = \frac{y^2}{4} \]

Now, substituting \( y = 2t \) and \( x = t^2 \) into the polar coordinates:
1. From \( y = 2t \):
\[ t = \frac{y}{2} = \frac{r \sin \theta}{2} \]

2. Substitute \( t \) into the equation for \( x \):
\[ x = t^2 = \left(\frac{r \sin \theta}{2}\right)^2 = \frac{r^2 \sin^2 \theta}{4} \]

Now, we also have:
\[ x = r \cos \theta \]

Equating both expressions for \( x \):
\[ r \cos \theta = \frac{r^2 \sin^2 \theta}{4} \]

To eliminate \( r \) (assuming \( r \neq 0 \)), we can rearrange this equation:
\[ 4r \cos \theta = r^2 \sin^2 \theta \]

Dividing both sides by \( r \):
\[ 4 \cos \theta = r \sin^2 \theta \]

Thus, we get the polar equation:
\[ r = \frac{4 \cos \theta}{\sin^2 \theta} \]

Now, using the identity \( \sin^2 \theta = \frac{1}{\csc^2 \theta} \), we can rewrite it as:
\[ r = 4 \cos \theta \csc^2 \theta \]

Then we can express it in terms of cotangent:
\[ r = 4 \cot \theta \csc \theta \]

This form matches the first option:

**Final Polar Equation:**

\[ r = 4 \cot \theta \csc \theta \]

So the correct choice is:
**r = 4 cot θ csc θ**