1. ## Polar Equation Help

Find the polar equation that has the same graph as the given Cartesian.

x^2-12y-36=0

I can't figure out how to do this problem.

2. Originally Posted by dachosin1
Find the polar equation that has the same graph as the given Cartesian.

x^2-12y-36=0

I can't figure out how to do this problem.
Start by substituting $x = r \cos \theta$ and $y = r \sin \theta$.

Show your all working and say where you get stuck.

3. x^2-12y-36=0

(rcosx)^2-12rsinx-36=0

(r^2)(cos(x)^2)-12rsinx-36=0

(r^2)(cos(x)^2)-12rsinx=36

rcos(x)^2-12sinx=36/r

rcos(x)^2=(36/r) + 12sinx

rcos(x)^2=(36/sqrt(x^2+y^2))+12sinx

r=36/[sqrt(x^2+y^2)(cos(x)^2)]+12sinx/cos(x)^2

is this the last step?

4. Originally Posted by dachosin1
x^2-12y-36=0

(rcosx)^2-12rsinx-36=0

(r^2)(cos(x)^2)-12rsinx-36=0

(r^2)(cos(x)^2)-12rsinx=36

rcos(x)^2-12sinx=36/r

rcos(x)^2=(36/r) + 12sinx

rcos(x)^2=(36/sqrt(x^2+y^2))+12sinx

r=36/[sqrt(x^2+y^2)(cos(x)^2)]+12sinx/cos(x)^2

is this the last step?
You are meant to get r in terms of $\theta$. That means no y's or x's.

You have $(\cos^2 \theta) r^2 - (12 \sin \theta) r - 36 = 0$. This is a quadratic equation in $r$. Solve for $r$ using the quadratic formula.

5. Perhaps we don't need to solve it by quadratic formula , juz take square root

x^2 - 12y - 36 = 0 <=>
r^2(cos^2 a) - 12r(sina) - 36 = 0 <=>
r^2[ 1 -(sin^2a)] - 12r(sina) - 36 = 0 <=>
r^2 - { [rsin(a)]^2 + 2*6[rsin(a)] + 6^2 } = 0 <=>
r^2 = [rsin(a) + 6]^2 <=>
r = rsin(a) + 6 OR r = -rsin(a) + 6
r = 6/[1 - sin(a)] OR r = 6/[ 1+ sin(a)]