How to Convert a Polar Equation to Cartesian Form?

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Find the Cartesian Form of

r(1+sinθ)=a
I understand that:
x=rcosθ
y=rsinθ
and r2=x2+y2I can get quite far to the point I end up with

x2=a2-2ay

I'm struggling to eliminate "a"...
 
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qablos said:
Find the Cartesian Form of

r(1+sinθ)=a



I understand that:
x=rcosθ
y=rsinθ
and r2=x2+y2


I can get quite far to the point I end up with

x2=a2-2ay

I'm struggling to eliminate "a"...

You can't eliminate a. I think you are already done.
 
Dick said:
You can't eliminate a. I think you are already done.

Okay...I've expressed my answer as:

y=1/2[a-(x2/a)]

Is that a better way to express it?
 
I feel it's slightly clearer as (a2-x2)/2a, but your answer is fine.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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