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How can I solve this form of quadratic? It has no $\sin(t)\cos(t)$ term.

$(\cos(t) + p + a)^2 - a^2 + b (\sin(t) + q)^2 = 0$

Multiplied out:

$\cos^2(t) + 2(a+p)\cos(t) + b\sin^2(t) + 2bq\sin(t) + (p^2 + 2ap + bq^2) = 0$

I'm at a loss for anything short of writing it out in complex exponentials. Is there another technique?

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    no the simple $\cos$ term will keep the sign (and you should double check anyway as you'll need to square out the root)2011-10-13

2 Answers 2

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Might I suggest the t-formulae?

$t = \tan\dfrac{x}{2}$

$\sin\,x = \dfrac{2t}{1+t^2}$

$\cos\,x = \dfrac{1-t^2}{1+t^2}$

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    More formally, this is called the Weierstrass substitution.2011-10-13
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Well, $(x + p + a)^2 - a^2 + b (y + q)^2 = 0$ is an ellipse (or hyperbola). Also, $(\cos t, \sin t)$ parametrizes a circle $x^2+y^2=1$. You want the intersection, which could have 4 points on it.