To eliminate the parameter [tex]\( t \)[/tex] from the parametric equations [tex]\( x = \cos(t) - 5 \)[/tex] and [tex]\( y = 3\sin(t) + 6 \)[/tex], we'll follow these steps:
1. Express the trigonometric functions in terms of [tex]\( t \)[/tex]:
[tex]\[
x = \cos(t) - 5
\][/tex]
[tex]\[
y = 3\sin(t) + 6
\][/tex]
2. Solve [tex]\( x = \cos(t) - 5 \)[/tex] for [tex]\( \cos(t) \)[/tex]:
[tex]\[
x + 5 = \cos(t)
\][/tex]
3. Next, isolate [tex]\( \cos(t) \)[/tex]:
[tex]\[
\cos(t) = x + 5
\][/tex]
4. Now, we need to relate [tex]\( y \)[/tex] to [tex]\( \sin(t) \)[/tex]. Given [tex]\( y = 3\sin(t) + 6 \)[/tex]:
[tex]\[
y - 6 = 3\sin(t)
\][/tex]
[tex]\[
\sin(t) = \frac{y - 6}{3}
\][/tex]
5. Recall the Pythagorean identity [tex]\( \sin^2(t) + \cos^2(t) = 1 \)[/tex]:
Using the expressions we derived for [tex]\( \cos(t) \)[/tex] and [tex]\( \sin(t) \)[/tex]:
[tex]\[
\left( \cos(t) \right)^2 + \left( \sin(t) \right)^2 = 1
\][/tex]
6. Substitute [tex]\( \cos(t) = x + 5 \)[/tex] and [tex]\( \sin(t) = \frac{y - 6}{3} \)[/tex] into the identity:
[tex]\[
(x + 5)^2 + \left( \frac{y - 6}{3} \right)^2 = 1
\][/tex]
7. Expand and simplify the equation:
[tex]\[
(x + 5)^2 + \left( \frac{y - 6}{3} \right)^2 = 1
\][/tex]
[tex]\[
(x + 5)^2 + \frac{(y - 6)^2}{9} = 1
\][/tex]
8. This equation is in the standard form of an ellipse:
[tex]\[
\frac{(x + 5)^2}{1} + \frac{(y - 6)^2}{9} = 1
\][/tex]
Therefore, the rectangular equation describing this curve is an equation of an ellipse:
[tex]\[
\frac{(x + 5)^2}{1} + \frac{(y - 6)^2}{9} = 1
\][/tex]
Conclusion:
The correct answer is that the rectangular equation represents an ellipse.
