Question

How do you write x=2sec(t) and y=3tan(t) in rectangular form?

Answer 1

`x^2/4 - y^2/9 = 1`

Explanation:

I want to be very lazy and use trig identities (instead of some form of inverse functions). The one that pops to mind is
`sec^2(t) = tan^2(t) + 1`
Solving for each of these factors in the given equations yields
`x/2 = sec(t) and y/3 = tan(t)`
hence
`(x/2)^2 = (y/3)^2 + 1 implies x^2/4 - y^2/9 = 1`
This is an hyperbola with dimension 2 in the x and 3 in the y.