we want
(x-h)^2/a^2 +(y-k)^2/b^2 = 1
now
x-h = a cos t
so
(x-h)^2 = a^2cos^2 t
so
(x-h)^2/a^2 = cos^2 t
similarly
(y-k)^2/b^2 = sin^2 t
so
(x-h)^2/a^2 + (y-k)^2/b^2 = cos^2 + sin^2 = 1
Can someone please help me with this question?
Show that
x=a cos t+h, y=b sin t+k; 0<t<2pi
(greater than and EQUAL to)
are parametric equations of an ellipse with center (h,k) and axes of lengths 2a and 2b.
Thank you so much!
1 answer
1 answer