finding rectangular equations from parametric equations

Question

how do you find the rectangular equation for the plane curve defined by the parametric equations: x=t-3 and y=t^2+5?

Answer 1

Hi Morgan,

Converting parametric equation to a cartesian equation or rectangular form involves solving for t in terms of x and then plugging this into the y equation.  Therefore,

t=x+3

y = (x+3)^2 + 5

y = x^2+6x + 9 + 5

y= x^2 + 6x + 14

 

Hope that helps.  Jim

Answer 2

x=t-3

t = x+3

y = t^2 + 5

y = (x+3)^2 +5

It's an upward opening parabola with vertex (-3,5) = minimum point, axis of symmetry is x=-3

y=x^2 +6x + 9 +5 = x^2 +6x +14

y intercept is (0,14)

the discriminant <0 so there are no x intercepts, just imaginary zeros

Answer 3

Yes, Jim is entirely correct. 

I would only augment his answer by pointing out that in general you need to manipulate the parametric equations to eliminate the parametric variable.  

It made the most sense to eliminate t by solving the first equation for t, in this particular problem, because that equation was linear so each value of x only corresponded to one value of t.  Things would have gotten a bit messy in this case if you had tried to solve the second equation for t, in this problem, but in general you eliminate the non rectangular-grid parameter by whatever means possible if you want the expression converted to its rectilinear coordinates. 

Which equation should be solved for the parametric variable depends on the problem -- whichever equation can be most easily solved for that parametric variable is typically the best choice.  For instance had the problem been y = t -3, and x = t^2 + 5, I hope you see that solving for t in terms of y would make more sense, for exactly the same reasons already discussed. 

So you solve for t in terms of x if that's easier, or solve for t in terms of y if that's easier, then put the result into the other equation. In this problem, the former was easier.