parabola help = =" (1 Viewer)

popoh8u

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i really dont get parametric equations of the parabola. .. = ="
doesnt anyone hv a summary of that a = ="
 
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pLuvia

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Post questions up and I'm sure people will help you.

Look through your textbooks, read the examples
 

Wackedupwacko

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basically its writing the caretesian equation (eg x² = 4ay) in a form where y = ? and x =? that will satisfy your cartesian equation. to turn it back all you hafta do is spot say the same variable in it eg x= 2at y=at² , make t the subject in the first one so u get
t= x/2a , sub into 2nd one and u get y= ax²/4a² , which is x² = 4ay

there are other forms but basically all you do is look to eliminate the variable that wont be used and maintaining your x and y...

another example is circle x = rcos@ and y= rsin@ : so you ask yourself how do u eliminate the cos and sin (as they obviously wont be in the cartesian form) well simple by adding the squares thus

x² = r²cos²@ and y²=r²sin²@ , add them together and u have:
x² + y² = r²(cos²@ + sin²@) <---- = 1 thus u have your basic circle
 

Riviet

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One of the hardest bits of parametrics is finding the locus. To find locus:
1. Find the general point of the locus, this will always be expressed in terms of a few variables, usually a, and 1 or 2 parameters, like p and q
2. Since this point is variable we let the co-ordinates be equations x=#$@#$ and y=@#$@#@$ where @#$@#% is some random expression, lol.
3. The locus will always be found by simultaneously solving the two equations, and will nearly always involve eliminating all the parameters, ending up with a cartesian equation (with only x, y and frequently a). This will be your locus. To eliminate the parameters, use a restriction on the parameters, e.g pq=-1 or p+q=2. This restriction will nearly always be asked to show earlier on in the question.
Finding the locus questions are nearly always at the end of a Q, and the Q will usually take you through everything like showing this and that, and finding a restriction, finally finding the locus.
 

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