Center of a parametric curve

[Dumbass]

Greetings!

What is the center of the conic of parametric equations:


Here is what i did:
i tried to found the cartesian equation of the given conic and then i used partial derivatives to find the center of the conic.
Here is what i did:



Is C(1,1) the center of the conic?

Thank you!

 

[Dr.Peterson]

You made a mistake right at the start, changing



to



Also, I wouldn't use inverse trig functions, which restrict the domain of x and y. Instead, I'd just solve for sine and cosine of 2t and use the Pythagorean identity. That also saves much of the second page.

But after you fix this, please define the "center" of a curve. What is it that you are finding on the last page, and how do you know that your work does that?

 

[Dumbass]

I have

You made a mistake right at the start, changing

View attachment 33076

to

View attachment 33077

Also, I wouldn't use inverse trig functions, which restrict the domain of x and y. Instead, I'd just solve for sine and cosine of 2t and use the Pythagorean identity. That also saves much of the second page.

But after you fix this, please define the "center" of a curve. What is it that you are finding on the last page, and how do you know that your work does that?

I fixed this i found that the center was (1,2).

The center of a conic section is the point midway between the foci.

We learned in school to find the center of a conic section using partial derivatives of their formula in cartesian form.

This is why i found the cartesian equation of the given conic.

 

[Dr.Peterson]

I have

I fixed this i found that the center was (1,2).

The center of a conic section is the point midway between the foci.

We learned in school to find the center of a conic section using partial derivatives of their formula in cartesian form.

This is why i found the cartesian equation of the given conic.

That's correct.

I'm not familiar with that method for finding the center, though I can see that it works, when the equation is in the right form. Having found that form, I would just complete the square and get more details about it: 4(x-1)^2+(y-2)^2=4.

 

[Dumbass]

That's correct.

I'm not familiar with that method for finding the center, though I can see that it works, when the equation is in the right form. Having found that form, I would just complete the square and get more details about it: 4(x-1)^2+(y-2)^2=4.

Thank you