+69 Find the Cartesian equivalent to the parametric equation x(t)= 3-2t and y(t)=(t^2) -2t
+26393 Find the Cartesian equivalent to the parametric equation x(t)= 3-2t and y(t)=(t^2) -2t
\(\begin{array}{|rcll|} \hline x(t)&=& 3-2t \\ x-3 &=& 2t \\ \mathbf{2t} &\mathbf{=}& \mathbf{ x-3 }\\ \mathbf{t} &\mathbf{=}& \mathbf{\frac12 (x-3) } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline y(t)&=&(t^2) -2t \qquad &| \qquad \mathbf{t =\frac12 (x-3) } \\ y &=&[ \frac12 (x-3) ]^2 -(2t) \qquad &| \qquad \mathbf{2t = x-3 }\\ y &=&[ \frac12 (x-3) ]^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3)\cdot \frac44 \\ y &=&\frac14 (x-3)\cdot [(x-3)-4] \\ y &=&\frac14 (x-3)\cdot [ x-3 -4] \\ y &=&\frac14 (x-3)\cdot ( x-7 ) \\ y &=&\frac14 (x^2-7x-3x+21) \\ \mathbf{y} &\mathbf{=}&\mathbf{ \frac14 \cdot (x^2-10x+21) }\\ \hline \end{array}\)
+26393 Find the Cartesian equivalent to the parametric equation x(t)= 3-2t and y(t)=(t^2) -2t
\(\begin{array}{|rcll|} \hline x(t)&=& 3-2t \\ x-3 &=& 2t \\ \mathbf{2t} &\mathbf{=}& \mathbf{ x-3 }\\ \mathbf{t} &\mathbf{=}& \mathbf{\frac12 (x-3) } \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline y(t)&=&(t^2) -2t \qquad &| \qquad \mathbf{t =\frac12 (x-3) } \\ y &=&[ \frac12 (x-3) ]^2 -(2t) \qquad &| \qquad \mathbf{2t = x-3 }\\ y &=&[ \frac12 (x-3) ]^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3) \\ y &=&\frac14 (x-3)^2 -(x-3)\cdot \frac44 \\ y &=&\frac14 (x-3)\cdot [(x-3)-4] \\ y &=&\frac14 (x-3)\cdot [ x-3 -4] \\ y &=&\frac14 (x-3)\cdot ( x-7 ) \\ y &=&\frac14 (x^2-7x-3x+21) \\ \mathbf{y} &\mathbf{=}&\mathbf{ \frac14 \cdot (x^2-10x+21) }\\ \hline \end{array}\)
