The rectangular equation of the plane curve, with parametric equations $x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,$ is

A) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 1$
B) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$
C) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$
D) $- \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$
E) $16 x ^ { 2 } + 9 y ^ { 2 } = 1$

Question

The rectangular equation of the plane curve, with parametric equations $x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,$ is

A) $\frac { x ^ { 2 } } { 9 } + \frac { y ^ { 2 } } { 16 } = 1$
B) $\frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$
C) $\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 } = 1$
D) $- \frac { x ^ { 2 } } { 16 } + \frac { y ^ { 2 } } { 9 } = 1$
E) $16 x ^ { 2 } + 9 y ^ { 2 } = 1$

Quizplus · Accepted Answer

To convert the parametric equations to a rectangular equation, we use the trigonometric identity $\sin^2 t + \cos^2 t = 1$. Given $x(t) = 4\sin t$ and $y(t) = 3\cos t$, we solve for $\sin t$ and $\cos t$, getting $\sin t = \frac{x}{4}$ and $\cos t = \frac{y}{3}$. Substituting these into the identity gives $\left(\frac{x}{4}\right)^2 + \left(\frac{y}{3}\right)^2 = 1$, which simplifies to $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

The Rectangular Equation of the Plane Curve, with Parametric Equations x(t)=4sin⁡t,y(t)=3cos⁡t,x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,x(t)=4sint,y(t)=3cost,

The Rectangular Equation of the Plane Curve, with Parametric Equations $x ( t ) = 4 \sin t , y ( t ) = 3 \cos t,$