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:
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From \( y = 2t \): \[ t = \frac{y}{2} = \frac{r \sin \theta}{2} \]
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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