Cartesian equation help?

Question

Choose the cartesian equation for the parametrized curve described by x = 4t, y = 16t62, -∞ ≤ t ≤ ∞. Describe the portion of the graph of the Cartesian equation that is traced by the parameterized curve.

a) y = x^2

b) y = 2x^2

c) y = 2^2

d) y = 4x

Any help on this would be appreciated. Please explain completely so that I fully understand and can do similar problems on my own. I have no base knowledge of this, so the more detailed the better.Thanks so much and 10 points to a best answer.

Answer 1

Solve one of the variables for the parameter, then put that into the other equation.

x = 4t ⇒ t = x/4

y = 16t² = 16(x/4)² = x²

so the answer is (a).

If -∞ ≤ t ≤ ∞, then x = 4t also runs from -∞ to ∞, so the entire parabola is traced by the given parametrization. If the limits on t had been 0 ≤ t ≤ ∞, then only the portion of the parabola on and to the right of the y-axis (that is, for x ≥ 0) would be traced.