# Math Help - Writing an ellipse in polar form a+ r cos(theta) = br.

1. ## Writing an ellipse in polar form a+ r cos(theta) = br.

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).

2. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

Have you at least attempted substituting \displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*} and \displaystyle \begin{align*} y = r\sin{(\theta)} \end{align*}?

3. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

Yes, I got to 8r^2 cos^2(theta) -16rcos(theta) +9r^2 sin^2(theta) = 64, but i'm not sure how to get this into the correct form.

4. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

Now notice that

\displaystyle \begin{align*} 8r^2 \cos^2{(\theta)} + 9r^2 \sin^2{(\theta)} &= 8r^2 \cos^2{(\theta)} + 8r^2 \sin^2{(\theta)} + r^2\sin^2{(\theta)} \\ &= 8r^2 \left[ \cos^2{(\theta)} + \sin^2{(\theta)} \right] +r^2 \sin^2{(\theta)} \end{align*}

Does this look more manageable now?

5. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

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?

6. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

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

7. ## Re: Writing an ellipse in polar form a+ r cos(theta) = br.

Got it now, thank you both