find the cartesian from this parametric??????hard one.?

Question

x = 1.5(t + 1/t)
y = 2.5(t-1/t)

no, we need to get the cartesian equation.
that is find y as a function of x.
example
if we had x = cost and y = sint
that is easy, x^2 + y^2 = 1 gets rid of parameter.

well done usp. thanx a lot
i will vote for u as soon as i can.

i guess its not that hard when you see the answer.
tnx both of u.

Answer 1

rearranging ur eqns

x/1.5 = t + 1/t

y/2.5 = t - 1/t


adding these eqns

x/1.5 + y/2.5 = 2t

or t = x/3 + y/5

subtracting the eqns

x/1.5 - y/2.5 = 2/t

or 1/t = x/3 - y/5

using the obtained eqns

x/3 + y/5 = 1 / ( x/3 - y/5)

or (x/3 +y/5 ) (x/3 -y/5 ) = 1

or x^2 / 9 - y^2 / 25 = 1

Answer 2

Your equations give you the Cartesian coordinates directly. Just substitute a value of t. For example, at t = 1,

x = 1.5 × 2 = 3
y = 2.5 × 0 = 0

so the Cartesian coordinates are (3,0).

Perhaps you are missing something from the question.

Answer 3

x = 1.5(t + 1/t)
y = 2.5(t-1/t)

So, 2x/3 =(t + 1/t)
And, 2y/5 = (t-1/t)
t = x/3 + y/5

So, 2x/3 = (5x+3y)/15 +15/(5x+3y)
10x(5x+3y) = (5x+3y)^2 +15^2
50x^2 +30 xy = 25x^2 + 30xy + 9y^2 +225
25x^2 - 9y^2 = 225

Answer 4

Well, I reckon t=5/(0.6x-y), so that should be it to work out the rest

Ashley