Polar equation to cartesian conversion problems

[rir0302]

1. xy = 19

So I did:
r²cos(theta)sin(theta) = 19
r²cos(theta)sin(theta) - 19 = 0
And because the question wants you to use the trig. identity
r²(1/2)(sin(2theta)) - 19 = 0

Is -19 wrong here? Or what did I do wrong?

The other questions are:
2. 1/(8-cos(theta))
3. 1 + 7cos(theta)
But I want to figure out #1 first

 

[Steven G]

Why do you think your work is wrong? Personally if I was going to use the double angle theorem, as you did, I would now multiply both sides by 2. I would even leave the 19 on the rhs.


And because the question wants you to use the trig. identity --Don't you think that it would be nice of you to show us the entire problem so we can help you?

 

[rir0302]

It's because it was an online question - it gave this as a hint when I got it wrong:

So I'm kind of confused now

 

[Cubist]

You have done nothing wrong so far. (Except you didn't need to move the 19, like Jomo said, so just move it back to the RHS)

The title you wrote might be confusing you, because you're actually aiming towards a polar equation. Do you remember what polar equations look like? If you do, then just keep going.

 

[rir0302]

Is this form correct? It's still marking it wrong.

 

[Dr.Peterson]

Why not do the same (solving for r), but with the [MATH]\sin(2\theta)[/MATH] you had? It does seem to be demanding that.

 

[Steven G]

View attachment 15218

Is this form correct? It's still marking it wrong.

I suggested that you convert r^2(1/2)(sin(2theta)) - 19 = 0 to r^2(sin(2theta))= 38