Original post by subbhy

Hi, how would I find the Cartesian form of this curve?

I rearranged f(t) for t but this results in square rooting and so part of the curve might be lost

The answer is 8y^2 = x^3 + 2x^2

I rearranged f(t) for t but this results in square rooting and so part of the curve might be lost

The answer is 8y^2 = x^3 + 2x^2

t^2-1 looks a common factor? so y^2 = ...

(edited 4 months ago)

Original post by mqb2766

t^2-1 looks a common factor? so y^2 = ...

Thanks

Original post by BraeLynnT

Yurp ion understand any of that. 🤷

Haha parametric curves are written implicitly in terms of another variable (the parameter). In this case the parameter is t. To get the Cartesian form of the equation (in terms of x and y only) we need to eliminate the parameter t. In this case squaring y to get terms of t^6, t^4 and t^2 allow us to make a substitution which is in terms of x. To get this substitution you rearrange x = 2t^2 - 2 to make t^2 the subject

(edited 4 months ago)

Original post by subbhy

Haha parametric curves are written implicitly in terms of another variable (the parameter). In this case the parameter is t. To get the Cartesian form of the equation (in terms of x and y only) we need to eliminate the parameter t. In this case squaring y to get terms of t^6, t^4 and t^2 allow us to make a substitution which is in terms of x. To get this substitution you rearrange x = 2t^2 - 2 to make t^2 the subject

From their history, Im not sure the previous post was serious, but it would be slightly simpler here to use the t^2-1 factor directly so

y = t(x/2)

then square and sub for t^2.

Original post by mqb2766

From their history, Im not sure the previous post was serious, but it would be slightly simpler here to use the t^2-1 factor directly so

y = t(x/2)

then square and sub for t^2.

y = t(x/2)

then square and sub for t^2.

Alright cool

Original post by subbhy

Haha parametric curves are written implicitly in terms of another variable (the parameter). In this case the parameter is t. To get the Cartesian form of the equation (in terms of x and y only) we need to eliminate the parameter t. In this case squaring y to get terms of t^6, t^4 and t^2 allow us to make a substitution which is in terms of x. To get this substitution you rearrange x = 2t^2 - 2 to make t^2 the subject

I barely understand geometry. Ainoway Ima be understanding that lol

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