Question

Here is a nasty rectangular to polar conversion, 4x^2-5y^2-36y-36=0...can anyone help solve for r? Thanks!

Answer 1

I got it down to this. I just can't solve the quadratic! \[(4 \cos^2 \theta -5 \sin^2 \theta) r^2 -(36 \sin \theta) r - 36 =0\]

Answer 2

@abb0t

Answer 3

@primeralph

Answer 4

[36sin(t)pm24sin(t)+24cos(t)] / [8cos^2(t)-10sin^2(t)]

Answer 5

@phi

Answer 6

@Hailey553

Answer 7

@kaylalynn

Answer 8

Since your equation:\[(4 \cos^2 \theta -5 \sin^2 \theta) r^2 -(36 \sin \theta) r - 36 =0\]has a \(\sin(\theta)\) in it, I would first replace the \(\cos^2(\theta)\) with \((1-\sin^2(\theta))\). You should then find that it simplifies quite a bit and, if you then use the quadratic equation to solve for \(r\), the \(\sin^2(\theta)\) terms inside the square root should cancel out to leave you with a number that is a perfect square.

Answer 9

thank you very much :) how do i give you a trophy?

Answer 10

just gave you one

Answer 11

thx - and you are more than welcome my friend! :)

Answer 12

I was looking for trig identities all over the web and i couldn't find the cos theta = 1 - sin^2theta. Thanks to you though i got it right! Calc 2 finals is pretty hefty haha

Answer 13

You might find this page useful: http://en.wikipedia.org/wiki/List_of_trigonometric_identities

Answer 14

perfect.