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If the graph of the equation $Ax^2+Cy^2+Dx+Ey+F=0$ is the empty set (no point at all), is it considered a conic? Isn't a conic supposed to be the intersection of a plane and a cone?

set5
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    There is no universal rule. People approaching conics from the equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ probably regard $x^2+y^2+1=0$ as a degenerate conic. People starting from a geometrical viewpoint probably do not. – almagest Dec 31 '19 at 18:38
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    @almagest That's just a circle of radius $i$ – Peter Foreman Dec 31 '19 at 18:40
  • True, but I assume the OP is only interested in real points. – almagest Dec 31 '19 at 18:42
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    A "conic" or "conic section" is the intersection of a plane with a (double) cone. Any plane at all will intersect such a cone in at least one point. So, no, the empty set is not a "conic". – user247327 Dec 31 '19 at 19:52

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The graph is empty only if you're dealing with strictly real coordinates. Otherwise there's always a solution by the fundamental theorem of algebra. So you see, it all depends on context ($\mathrm R$ or $\mathrm C$) whether you consider all such equations as representing conics, or only those having real points.

Allawonder
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