Have you at least attempted substituting and ?
The given ellipse is (x-1)^2 /9 + y^2 /8 = 1
Show that the equation of the given ellipse in polar coordinates has the form
a + r cos(theta) = b r and find the values of a and b.
This is an ellipse with x-intercepts at -2 and 4, and centre at (1,0).
Ok, so I reduced that to 8r^2 + r^2(sin^2 theta) = 64
then taking square roots and rearranging,
sqrt(8)r = -8 + rsin(theta)
But this is in terms of sin and not cos, like the question says. Can this be solved by letting x= rsin(theta) and y = rcos(theta) [the opposite substitution as before] so that the final answer is in terms of cos(theta)?
[EDIT] I forgot about the extra -16rcos(theta) term.
So, using x=rcos(theta) and y = rsin(theta) this gives
8r^2 + r^2 sin^2 (theta) -16rcos(theta) = 64. But how can I get rid of that extra rcos(theta) term?
8r^2 + r^2 sin^2 (theta) -16rcos(theta) = 64.
8r^2 + r^2 ( 1-cos^2 (theta)) -16rcos(theta) = 64.
8r^2 + r^2 - r^2 cos^2 (theta) -16rcos(theta) = 64.
9r^2 - r^2 cos^2 (theta) -16rcos(theta) = 64.
9r^2 - 64 = r^2 cos^2 (theta) +16rcos(theta)
9r^2 - 64 = r^2 cos^2 (theta) +16rcos(theta) + 64 - 64
9r^2 = [ rcos (theta) +8 ]^2
Now you can take it further