Answer:
The rectangular equivalence is:
[tex](\frac{x-2}{4})^2+(\frac{y+1}{5})^2=1[/tex]
The interval of x is [-2, 6]
Explanation:
To find the rectangular equivalence, we want an equation of the form:
[tex]\cos^2(\theta)+\sin^2(\theta)=1[/tex]
Then, we have the equations:
[tex]\begin{gathered} x(\theta)=4\cos(\theta)+2 \\ y(\theta)=5\sin(\theta)-1 \end{gathered}[/tex]
On each equation, we solve for cos and sin:
[tex]\frac{x-2}{4}=\cos(\theta)[/tex][tex]\frac{x-2}{4}=\cos(\theta)[/tex][tex]\frac{y+1}{5}=\sin(\theta)[/tex]
Now we can square both sides:
[tex]\begin{gathered} (\frac{x-2}{4})^2=\cos^2(\theta) \\ \end{gathered}[/tex][tex](\frac{y+1}{5})^2=\sin^2(\theta)[/tex]
Now we can write:
[tex](\frac{x-2}{4})^2+(\frac{y+1}{5})^2=1[/tex]
That's the rectangular equivalence of the parametric equations.
Now, to find the interval where x falls under, we have:
[tex]x(\theta)=4\cos(\theta)+2[/tex]
In this function, the value of x depends only on of θ. The maximum value that cos(θ) is 1, when θ = 0
Then, if θ = 0, cos(θ) = 1
[tex]x(0)=4\cdot1+2=6[/tex]
The minimum value of cos(θ) is -1, when θ = π
If θ = π, cos(θ) = -1
Then:
[tex]x(\pi)=4\cdot(-1)+2=-2[/tex]
The interval is [-2, 6]
