Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve. $x = 1 + 3 \cos \theta$ $$y = 1 + 5 \sin \theta$$
A) $\frac { ( x - 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$
B) $$\frac { ( x - 1 ) ^ { 2 } } { 9 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$$
C) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$
D) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 1$
E) $\frac { ( x - 1 ) ^ { 2 } } { 25 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1$

Question

Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.​ $x = 1 + 3 \cos \theta$ $$y = 1 + 5 \sin \theta$$ ​
A) $\frac { ( x - 1 ) ^ { 2 } } { 9 } + \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$
B) $$\frac { ( x - 1 ) ^ { 2 } } { 9 } - \frac { ( y - 1 ) ^ { 2 } } { 25 } = 1$$
C) $\frac { x ^ { 2 } } { 9 } - \frac { y ^ { 2 } } { 25 } = 1$
D)​ $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 25 } = 1$
E) $\frac { ( x - 1 ) ^ { 2 } } { 25 } - \frac { ( y - 1 ) ^ { 2 } } { 9 } = 1$

Quizplus · Accepted Answer

From the given parametric equations $x = 1 + 3 \cos \theta$ and $y = 1 + 5 \sin \theta$, we can isolate $\cos \theta$ and $\sin \theta$ as $\cos \theta = \frac{x - 1}{3}$ and $\sin \theta = \frac{y - 1}{5}$. Using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we substitute these expressions and simplify to get $\frac{(x - 1)^2}{9} + \frac{(y - 1)^2}{25} = 1$, which matches choice A.

Eliminate the Parameter and Write the Corresponding Rectangular Equation Whose x=1+3cos⁡θx = 1 + 3 \cos \thetax=1+3cosθ

Eliminate the Parameter and Write the Corresponding Rectangular Equation Whose $x = 1 + 3 \cos \theta$