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Mathematics 18 Online
OpenStudy (anonymous):

Eliminate the parameter t AND write the rectangular equation for x = t2+3 and y = 2t.

OpenStudy (anonymous):

CORRECTION: x= t^2+3

OpenStudy (anonymous):

t=y/2

OpenStudy (anonymous):

x= y^2/4 +3

OpenStudy (anonymous):

I'm confused....i got t = sqrt of 3-x.

OpenStudy (anonymous):

\[x=\frac{y^2}{4}+3\] 4 x = y^2 + 3 4x-3=y^2

OpenStudy (anonymous):

then you have to use plus or minus i think. that is why iman solved the other way. correct me if i am wrong

OpenStudy (anonymous):

can u explain how you got x = y^2/4 =3

OpenStudy (anonymous):

y=2t t=y/2 t^2 = y^2/4

OpenStudy (anonymous):

Oh i see...but what is a rectangular equation??

OpenStudy (anonymous):

where you equation is like y=f(x)

OpenStudy (anonymous):

as opposed to parametric equation where both x and y depend on t

OpenStudy (anonymous):

so is the solution to the question 4x-3=y^2 or something else

OpenStudy (anonymous):

y^2=4x-3 take square root of both side \[y=\pm \sqrt{4 x-3}\]

OpenStudy (anonymous):

ok...i get it now...thanks..is that the end of the problem

OpenStudy (anonymous):

yes, relieved?

OpenStudy (anonymous):

yup! the question was worded to weirdly...i wasnt evn sure what it was asking

OpenStudy (anonymous):

can you help me with anotehr q?

OpenStudy (anonymous):

sure , but post it as new question

OpenStudy (anonymous):

ok...ill post it now

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