Solving a Polar Equation with Negative Numerator: Is It Possible?

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I have the given polar equation that I have to graph.

r = 5/(-1 + 2 cos θ)

The format is r = ed/(1 +e cosθ)

(e and d are always positive)

I can solve multiply r by -1/-1 --> r = -5/(1 - 2 cosθ).

Is this possible? I thought that the numerator (ed) cannot be a negative number. Is this a misprint or is there a way to solve this?
 
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r can be negative... it just means that you actually go 180 degrees the other direction. So for example, when theta is 180 degrees, r is -5/3... so the point is at 5/3 on the *positive* x-axis.
 
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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