you know that
x=r cosθ
y=r sinθ
describes a circle of radius r, with center at (0,0).
If the axes are a and b, instead of r,
x = a cosθ
y = b sinθ
If the center is at (h,k) just translate the coordinates:
x = a cosθ + h
y = b sinθ + k
The graph of the equation (x-h)^2/(a^2) + (y-k)^2/b^2 = 1 is an ellipse with center (h,k), horizontal axis length 2a, and vertical axis length 2b. Find parametric equations whose graph is an ellipse with center (h,k), horizontal axis length 2a, and vertical axis length 2b, and explain why your answer is correct.
I really have no idea where to begin. I'm sorta new to the concept of parametric equations, so could someone please explain how to solve this? Thanks!
3 answers
(x-h)^2 /a^2 + (y-k)^2 /b^2 = 1
we know that sin^2 t + cos^2 t = 1
so let cos^2 t = (x-h)^2/a^2
cos t = (x-h)/a
and let sin^2 t = (y-k)^2 / b^2
sin t = (y-k)/b
from
(x-h)/a = cos t
x-h = acos t
x = h + a cos t
from
(y-k)/b = sin t
y-k = b sin t
y = k + b sint
thus:
x=h + acost
y = k + bsint
Here is a Youtube that might shed light on this
http://www.youtube.com/watch?v=zs0Nw0tb4y8
we know that sin^2 t + cos^2 t = 1
so let cos^2 t = (x-h)^2/a^2
cos t = (x-h)/a
and let sin^2 t = (y-k)^2 / b^2
sin t = (y-k)/b
from
(x-h)/a = cos t
x-h = acos t
x = h + a cos t
from
(y-k)/b = sin t
y-k = b sin t
y = k + b sint
thus:
x=h + acost
y = k + bsint
Here is a Youtube that might shed light on this
http://www.youtube.com/watch?v=zs0Nw0tb4y8
Thank you very much both of you sirs it is much appreciated!