How do you transform parametric equations into Cartesian form: x= 3 + 2 cost and y= 1 + 5sint?

Question

William Abdalla · Accepted Answer

#25x^2+4y^2-150x-8y+129=0#

As #x=3+2cost# and #y=1+5sint#,

we have #cost=(x-3)/2# and #sint=(y-1)/5#

Hence #((x-3)/2)^2+((y-1)/5)^2=1# or

#(x^2-6x+9)/4+(y^2-2y+1)/25=1# or

#25(x^2-6x+9)+4(y^2-2y+1)=100# or

#25x^2+4y^2-150x-8y+225+4-100=0# or

#25x^2+4y^2-150x-8y+129=0#

graph{25x^2+4y^2-150x-8y+129=0 [-7.295, 12.705, -3.92, 6.08]}

Evelyn Agard · Answer

undefined To transform parametric equations into Cartesian form, eliminate the parameter by solving for it in one of the equations and substituting it into the other equation. For $ x = 3 + 2\cos(t) $ and $ y = 1 + 5\sin(t) $, square both equations and rearrange to solve for $\cos(t)$ and $\sin(t)$ respectively. Then, substitute these expressions into one of the original equations to obtain the Cartesian equation.

$ x = 3 + 2\cos(t) $

$ x - 3 = 2\cos(t) $

$ \cos(t) = \frac{x - 3}{2} $

$ y = 1 + 5\sin(t) $

$ y - 1 = 5\sin(t) $

$ \sin(t) = \frac{y - 1}{5} $

Substitute $\cos(t) = \frac{x - 3}{2}$ into $ \sin^2(t) + \cos^2(t) = 1$ to eliminate the parameter.

$ \sin^2(t) + \left(\frac{x - 3}{2}\right)^2 = 1 $

$ \sin^2(t) + \frac{(x - 3)^2}{4} = 1 $

$ \sin^2(t) = 1 - \frac{(x - 3)^2}{4} $

$ \sin(t) = \pm \sqrt{1 - \frac{(x - 3)^2}{4}} $

Substitute $ \sin(t) = \frac{y - 1}{5} $ into the equation above.

$ \frac{y - 1}{5} = \pm \sqrt{1 - \frac{(x - 3)^2}{4}} $

Solve for $ y $.

$ y - 1 = \pm 5\sqrt{1 - \frac{(x - 3)^2}{4}} $

$ y = 1 \pm 5\sqrt{1 - \frac{(x - 3)^2}{4}} $