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C4 Question

Re: Edexcel C4 Book/ Review exercises section p138/Question 71/part c
So, as you can see, the request is for me to find a cartesian equation for the parametric equations x=tan^{2}t, y=sint

...so I start playing with the y=sint equation and I get:
y^{2}=sin^{2}t=1-cos^{2}t

y^{2}-1=-cos^{2}t
1-y^{2}=cos^{2}t

Now looking at:
x=tan^{2}t=\frac{sin^{2}t}{cos^{2}t}

Substituting the identities for y we get:
x=\frac{y^{2}}{1-y^{2}}
This, as you can see, is an improper fraction---after dividing out the improper fraction on RHS I get:
-1+\frac{1}{1-y^{2}}
Now....
x=-1+\frac{1}{1-y^{2}}
1-y^{2}=\frac{1}{x+1}
-y^{2}=-1+\frac{1}{x+1}
y^{2}=1-\frac{1}{x+1}
y^{2}=\frac{x+1-1}{x+1}

My question is: IS THIS WAY CORRECT?
(edited 4 years ago)
Original post by saizperez
Re: Edexcel C4 Book/ Review exercises section p138/Question 71/part c
So, as you can see, the request is for me to find a cartesian equation for the parametric equations x=tan^{2}t
, y=sint

...so I start playing with the y=sint equation and I get:
y^{2}=sint^{2}t=1-cos^{2}t

y^{2}-1=-cos^{2}t
1-y^{2}=cos^{2}t

Now looking at:
x=tan^{2}t=sin^{2}t/cos^{2}t

Substituting the identities for y we get:
x=y^{2}/1-y^{2}


You can just leave it in x=y21−y2x = \dfrac{y^2}{1-y^2} form... nothing wrong with that. You just might want to state the values which yy takes though so that the curves match perfectly.
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saizperez
OP
6
ok, but, fundamentally speaking, they wouldn't dock me marks right? besides half one for omitting the range.
Original post by RDKGames
You can just leave it in x=y21−y2x = \dfrac{y^2}{1-y^2} form... nothing wrong with that. You just might want to state the values which yy takes though so that the curves match perfectly.
(edited 4 years ago)
Original post by saizperez
ok, but, fundamentally speaking, they wouldn't dock me marks right? besides half one for omitting the range.


Doubt they would unless they ask for a specific form.

All a Cartesian eqn is its just an equation not involving the parameter... so you technically answer the question regardless whether you put x=... or y=...

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