Answer:
See below for answers and explanations
Step-by-step explanation:
Problem 1
[tex]x=-3+2cos\theta,\:y=5+2sin\theta\\\\x+3=2cos\theta,\: y-5=2sin\theta\\\\(x+3)^2=4cos^2\theta,\: (y-5)^2=4sin^2\theta\\\\(x+3)^2+(y-5)^2=4cos^2\theta+4sin^2\theta\\\\(x+3)^2+(y-5)^2=4(cos^2\theta+sin^2\theta)\\\\(x+3)^2+(y-5)^2=4(1)\\\\(x+3)^2+(y-5)^2=4[/tex]
Thus, the first option is correct. Trying all the other options will not get you the desired rectangular equation.
Problem 2
[tex]x=3-6cos\theta,\: y=-2+3sin\theta\\\\x-3=-6cos\theta,\: y+2=3sin\theta\\\\\frac{x-3}{-6}=cos\theta,\: \frac{y+2}{3}=sin\theta\\ \\ \frac{(x-3)^2}{36}=cos^2\theta,\: \frac{(y+2)^2}{9}=sin^2\theta\\ \\ \frac{(x-3)^2}{36}+\frac{(y+2)^2}{9}=cos^2\theta+sin^2\theta\\\\ \frac{(x-3)^2}{36}+\frac{(y+2)^2}{9}=1[/tex]
Therefore, the first option is correct. This equation is in the form of an ellipse with a horizontal major axis length of 12 (half is 6) and a vertical minor axis length of 6 (half is 3), with its center at (3,-2).
Problem 3
Not sure which equation needs to be used for this problem
Problem 4
[tex]x=-7cos\theta ,\:y=5sin\theta\\\\-\frac{x}{7}=cos\theta,\: \frac{y}{5}=sin\theta\\ \\ \frac{x^2}{49}=cos^2\theta,\: \frac{y^2}{25}=sin^2\theta\\ \\\frac{x^2}{49}+\frac{y^2}{25}=cos^2\theta+sin^2\theta\\ \\ \frac{x^2}{49}+\frac{y^2}{25}=1[/tex]
This equation is in the form of an ellipse with a horizontal major axis length of 14 (half is 7) and a vertical minor axis length of 10 (half is 5). See attached graph.
Problem 5
Eliminate the parameter:
[tex]x=-t^2-2,\:y=-t^3+4t\\\\x+2=-t^2\\\\-x-2=t^2\\\\\pm\sqrt{-x-2}=t\\\\y=-t^3+4t\\\\y=-(\pm\sqrt{-x-2})^3+4(\pm\sqrt{-x-2})[/tex]
Attached below is the graph of the curve, which corresponds with the first option.
