A curve has a vector equation
\(r=(2 \sin t-1)\mathbf{i}+(2 \cos t+3) \mathbf{j} \)
- Write the Cartesian equation of this curve. (2 marks)
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- Sketch the curve on the Cartesian plane below. (1 mark)
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Show Answers Only
a. \((x+1)^2+(y-3)^2=4\)
b. \((x+1)^2+(y-3)^2=2^2\)
\(\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2\)
Show Worked Solution
a. \(r=(2 \sin t-1) \mathbf{i}+(2 \cos t+3) \mathbf{j}\)
\(x=2 \sin t-1 \ \Rightarrow \ \sin t=\dfrac{x+1}{2}\)
\(y=2 \cos t+3 \ \Rightarrow \ \cos t=\dfrac{y-3}{2}\)
\(\text{Using} \ \ \sin ^2 t+\cos ^2 t=1:\)
| \(\dfrac{(x+1)^2}{4}+\dfrac{(y-3)^2}{4}\) | \(=1\) | |
| \((x+1)^2+(y-3)^2\) | \(=4\) |
b. \((x+1)^2+(y-3)^2=2^2\)
\(\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2\)