x=3sin t, y= 4cos t?

Question

eliminate the parameter

Answer 1

Hm, ok.

What do you think of this:

16x² + 9y² =
16(3sint)² + 9(4cost)² =
16(9)sin²t + 9(16)cos²t =
144sin²t + 144cos²t =
144(sin²t + cos²t) =
144

so 16x² + 9y² = 144

Answer 2

x/3 = sin t, y/4 = cos t

Square both sin t and cos t to get:

x^2/9 + y^2/16 = sin^2 t + cos^2 t = 1

Multiply both sides by the LCM of 9 and 16 to clear fractions.

Then we get:

16*x^2 + 9*y^2 = 144.

This is the equation of an ellipse, centered at (0,0) with major radius, R = 4, on the x-axis, and minor radius, r = 3, on the y-axis.

Answer 3

sin t = x/3 ----(1)
cos t = y/4 -----(2)

then we know that sin^2 t + cos^2 t = 1 , by Pythagoras' theorem.
then square (1) and (2) and equate to 1.
then,
(x^2 / 9) + (y^2 / 16) = 1

Answer 4

square both sides:
x^2=9sin^2t
y^2=16cos^2t
Divide the constants:
x^2/9
y^2/16
Add expressions:
X^2/9 + y^2/16 = 1
I'll assume you can do the rest. :)