Thread: Parametric Equations::

Find the Rectangular form of each pair of parametric Equations:

Hello, ^_^Engineer_Adam^_^!

Find the rectangular form of: . x .= .3·tan²θ . . y .= .2·sec²θ

. . - - - . . .x
We have: .-- .= .tan²θ . [1]
. . - - - . . .3

. . - - - . . y
. . . . . . . -- .= .sec²θ . [2]
. . - - - . . 2

. . . . . . . . . . . . . . . . .y . .x
Subtract [1] from [2]: . -- - -- .= .sec²θ - tan²θ .= .1
. . . . . . . . . . . . . . . . .2 . .3


We have the line: .y .= .(2/3)x + 2


But be careful!

Since tan²θ > 0, then x > 0

Since sec²θ > 1, then y > 2


The graph looks like this:

Code:

        |                   *
        |               *
        |           *
        |       *
        |   *
       2*
        |
        |
    - - + - - - - - - - - - - -
        |