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Equivalent condition of "$Ax^2+Bx+C=0$ has exactly one solution" is "$\Delta=B^2-4AC=0$".

Now we turn to the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$. Do we have discriminant for it? How to relate $A,B,...,F$ so that it has exactly one solution $(x,y)$? Thank you.

p.s. I searched wiki, by defining $\Delta=B^2-4AC=0$ again, $\Delta>0,=0,<0$ correspond to ellipse, circle, parabola, etc. But it says nothing about the above case.

JSCB
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  • Over the complex numbers, a nonconstant polynomial in two variables has infinitely many solutions (see here for more on this.) The question of whether there is only one (or are only finitely many) solutions $(x,y)$ with $x,y \in \mathbb R$ -- as in the case of the equation $x^2+y^2=0$ -- is more subtle. I do not know of a "discriminant-like" criterion. – vociferous_rutabaga Sep 07 '15 at 13:53
  • https://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections – Will Jagy Sep 07 '15 at 17:23

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You need an ellipse of dimensions zero, centered at $(p,q)$. Note I have $2B$ where you have $B$. $$A(x-p)^2+2B(x-p)(y-q)+C(y-q)^2=0\\B^2<AC\\ Ax^2+2Bxy+Cy^2-(2Ap+2Bq)x-(2Cq+2Bp)y+Ap^2+2Bpq+Cq^2=0\\ F=\left[\begin{array}{cc}p&q\end{array}\right] \left[\begin{array}{cc}A&B\\B&C\end{array}\right]\left[\begin{array}{c}p\\q\end{array}\right]= \left[\begin{array}{c}D&E\end{array}\right] \left[\begin{array}{cc}A&B\\B&C\end{array}\right]^{-1} \left[\begin{array}{c}D\\E\end{array}\right]/4$$

Empy2
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