Question

Eliminate the parameter t. Find a rectangular equation for the plane curve defined by the parametric equations.

x = 6 cos t, y = 6 sin t; 0 ≤ t ≤ 2π

A. x^2 - y^2 = 6; -6 ≤ x ≤ 6
B. x^2 - y^2 = 36; -6 ≤ x ≤ 6
C. x^2 + y^2 = 36; -6 ≤ x ≤ 6

Answer 1

For the answer to this question, I'll show you the two ways,


Take square
x^2 = 36cos^2t
y^2 = 36sin^2t , add both

x^2+y^2= 36( cos^2t+sin^2t)
x^2+y^2= 36 (1)
x^2+y^2=36 a circle with radius R^2=36 , ie, R=+-6 , so -6<x<6 ANSWER C

The other way ,
cos t = x/6 in a right triangle , adjacent side is x , hypotenuse is 6
so opposite side is b^2= 6^2-x^2
b^2 = 36-x^2
thus, sin t = b/6 ,
sint = sqrt ( 36-x^2) /6
We know that
y = 6sint

y = 6 sqrt (36-x^2) /6
y = sqrt (36-x^2)
taking square
x^2+y^2=36