if the parameter is eliminated from x=2Tan t and y=2sec t, what is the resulting equation?

Answer 1

Use the identity tan^2 + 1 = sec^2:

x^2 = 4 tan^2 t
y^2 = 4 sec^2 t = 4(tan^2 t + 1) = 4tan^2 t + 4 = x^2 + 4

So the cartesian equation is:

y^2 = x^2 + 4

Answer 2

Given the parametric equations:

x = 2tan t
y = 2sec t

Eliminate t and write the equation in terms of x and y only.

tan t = x/2
sec t = y/2

tan² t = (x/2)² = x²/4
sec² t = (y/2)² = y²/4

Recall the identity

sec² t = tan² t + 1

Then

y²/4 = x²/4 + 1
y²/4 - x²/4 = 1

This is a hyperbola centered at the origin that opens horizontally.

Answer 3

tan t =x/2
sec^2 t =1+tan^2 t
=1+(x/2)^2
sec t=sqrt(1+(x/2)^2)
now
y=2*sqrt(1+(x/2)^2)
sqrt is square root