Eliminate the parameter t?

Question

Eliminate the parameter t

x = 1 - 4cos(t) , y = -5 + 6sin(t)

I know that the answer is (x-1)^2 / 16 + (y+5)^2 / 36 = 1, but I don't know how or why that's the answer.

Answer 1

(x - 1) / 4 = cos t
(y + 5) / 6 = sin t

(x - 1)² / 16 + (y + 5)² / 16 = cos ² t + sin ² t
(x - 1)² / 16 + (y + 5)² / 16 = 1

Answer 2

x = 1 - 4cos(t) . . . . . cos t = (1 - x ) /4 . . . . cos ² t = (1 - x ) ² /16
y = -5 + 6sin(t) .. . . sin t = -(5+y)/6 . . . . . sin ² t = (5+y)² / 36
adding
sin ² t + cos ² t = (5+y)² / 36 + (1 - x ) ² /16 = 1
(5+y)² / 36 + (1 - x ) ² /16 = 1

Answer 3

Use the trigonometric identity sin^2 x + cos^2 x = 1.

If x = 1 - 4cos(t), solve for cos(t). So (x-1)/-4 = cos(t). Now for the y equation, solve for sin(t). y = -5 + 6sin(t). (y+5)/6 = sin(t).

Now square both of these.
(x-1)^2 / (-4)^2 = cos^2 (t).
(x-1)^2 / 16 = cos^2 (t).

(y+5)^2 / 6^2 = sin^2 (t).
(y+5)^2 / 36 = sin^2 (t).

Now plug into our trigonometric identity.
sin^2 (t) + cos^2 (t) = 1.
(y+5)^2 / 36 + (x-1)^2 / 16 = 1.

Answer 4

Write your equations in the following form!

(x-1)/4 = - cos t

(y+5)/6 = sin t

Square both sides and add them up, you'll get
the form you want:

(x-1)^2/16 + (y+5)^2/36 = cos^2 t + sin^2 t = 1

Answer 5

(x+1) = -4cos^2(t)
(x+1)^2=16cos^2(t)
((x+1)^2)/16=cos^2(t)

(y+5)=6sin(t)
(y+5)^2=36sin^2(t)
((y+5)^2)/36=sin^2(t)

Since sin^2(t)+cos^2(t)=1
((x+1)^2)/16 + ((y+5)^2)/36 = 1

Answer 6

write the first equation as cost=(1-x)/4
and the second as sint=(y+5)/6

square both sides of both equations, and get:

cos^2t+sin^2t= (1-x)^2/16+(y+5)^3/36

you know from trig that cos^2t+sin^t=1, so the left hand side becomes one, and that solves your equation (1-x)^2 is the same as (x-1)^2, so we are in agreement

Answer 7

x = 1 -- 4cost whence cost = (1--x)/4
y = --5 + 6sint whence sint = (y+5)/6
then (cost)^2 + (sint)^2 = 1
gives [(1--x)/4]^2 + [(y+5)/6]^2 = 1
Or (1--x)^2/16 + (y+5)^2/36 = 1