Answer:
[tex]\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1[/tex]
Step-by-step explanation:
[tex]x=2-3\cos(t)[/tex]
[tex]y=1+4\sin(t)[/tex]
Let's solve for [tex]\cos(t)[/tex] in the first equation and then solve for [tex]\sin(t)[/tex] in the second equation.
I will then use the following identity to get right of the parameter, [tex]t[/tex]:
[tex]\cos^2(t)+\sin^2(t)=1[/tex] (Pythagorean Identity).
Let's begin with [tex]x=2-3\cos(t)[/tex].
Subtract 2 on both sides:
[tex]x-2=-3\cos(t)[/tex]
Divide both sides by -3:
[tex]\frac{x-2}{-3}=\cos(t)[/tex]
Now time for the second equation, [tex]y=1+4\sin(t)[/tex].
Subtract 1 on both sides:
[tex]y-1=4\sin(t)[/tex]
Divide both sides by 4:
[tex]\frac{y-1}{4}=\sin(t)[/tex]
Now let's plug it into our Pythagorean Identity:
[tex]\cos^2(t)+\sin^2(t)=1[/tex]
[tex]\frac{x-2}{-3})^2+(\frac{y-1}{4})^2=1[/tex]
[tex]\frac{(x-2)^2}{(-3)^2}+\frac{(y-1)^2}{4^2}=1[/tex]
[tex]\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1[/tex]
