Inactive Tutor answered 11/06/23
x2 = 36 cos2 t, x2 / 36 = cos2 t
y2 = 9 sin2 t, y2 / 9 = sin2 t
(x2 / 36) + (y2 / 9) = sin2 t + cos2 t
(x2 / 36) + (y2 / 9) = 1
x2 + 4y2 = 36
Avril K.
asked 11/06/23Eliminate the parameter, t, in the parametric equations, x = 6cost and y = 3sint.
Inactive Tutor answered 11/06/23
x2 = 36 cos2 t, x2 / 36 = cos2 t
y2 = 9 sin2 t, y2 / 9 = sin2 t
(x2 / 36) + (y2 / 9) = sin2 t + cos2 t
(x2 / 36) + (y2 / 9) = 1
x2 + 4y2 = 36
Inactive Tutor answered 12/13/25
x=6cost
y=3sint
x^2= 36cos^2(t)
y^2 = 9sin^2(t)
x^2/36 = cos^2(t)
y^2/9 = sin^2(t)
x^2/36 + y^2/9 = 1
x^2 + 4y^2 = 36
Inactive Tutor answered 11/06/23
cos(t) = x/6
sin(t) = y/3
sin2(t) + cos2(t) = 1
(y/3)2 + (x/6)2 = 1
y2/9 + x2/36 = 1
x2/36 +4y2/36 = 1
x2+4y2= 36
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