The rectangular equation of the plane curve, with parametric equations $x ( t ) = e ^ { - t } , y ( t ) = 3 e ^ { t }$ with $t \in ( - \infty , \infty )$ , is

A) $y = \frac { 3 } { x }$
B) $y = \frac { x } { 3 }$
C) $y = - \frac { 3 } { x }$
D) $y = 3 x$
E) $y = - \frac { x } { 3 }$

Question

The rectangular equation of the plane curve, with parametric equations $x ( t ) = e ^ { - t } , y ( t ) = 3 e ^ { t }$ with $t \in ( - \infty , \infty )$ , is

A) $y = \frac { 3 } { x }$
B) $y = \frac { x } { 3 }$
C) $y = - \frac { 3 } { x }$
D) $y = 3 x$
E) $y = - \frac { x } { 3 }$

Quizplus · Accepted Answer

To eliminate the parameter $t$, we express both $x$ and $y$ in terms of $t$ and then solve for $t$ in one of the equations to substitute into the other. Given $x(t) = e^{-t}$ and $y(t) = 3e^t$, we can solve for $t$ in terms of $x$ by rearranging the first equation: $t = -\ln(x)$. Substituting $t$ from the first equation into the second gives us a direct relationship between $x$ and $y$. However, a simpler way is to notice that $x \cdot y = e^{-t} \cdot 3e^t = 3$, which simplifies to $y = \frac{3}{x}$, showing that the relationship between $x$ and $y$ is directly given by option A without explicitly solving for $t$.

The Rectangular Equation of the Plane Curve, with Parametric Equations x(t)=e−t,y(t)=3etx ( t ) = e ^ { - t } , y ( t ) = 3 e ^ { t }x(t)=e−t,y(t)=3et

The Rectangular Equation of the Plane Curve, with Parametric Equations $x ( t ) = e ^ { - t } , y ( t ) = 3 e ^ { t }$