What does partial differentiation give for a second degree equation which doesn't represent a conic?

Question

Question says: Find real solution to the equation $$3x^{2}+3y^{2}-4xy+10x-10y+10=0.$$ My first thought was to treat it as a general conic ( $ax^2 + 2hxy + by^2 + 2gx + 2fy + c=0 $) but when I do so, its discriminant is lesser than $0$. And on plotting on geogebra, also it only shows a point.

And this is the same point we get after doing partial differentiation and solving for $x,y$. Generally for a conic, this $(x,y)$ represents the center of the conic. But for equations like these which don't represent a conic, how does doing partial differentiation and then solving, gives us the integer solution to it. If it is center, then there should be a conic too.

My friends told that conic is in a complex plane, but shouldn't center should also be in complex plane? Sorry Idk much about conics which exist in both planes, so maybe what I wrote in last 2 lines is completely wrong. Pardon me for that.

Answer 1

A conic curve, is the intersection of the 3D cone $p_1z^2=p_2x^2+p_3y^2$ with the 3D plane $p_4x+p_5y+p_6=0$, which takes the general form as stated in the OP. Now, a point can also be regarded as a conic curve (or conic, simply) in a sense that it is the intersection of the 3D plane with the vertex of the 3D cone, i.e. the point in which, the two halves of the cone meet. An example is the intersection of $z^2=x^2+y^2$ with $2z=x+y$. Hence, your approach is correct. Another way for justifying the result, is to note that any single point is an ellipse with zero diameters and ellipse is a conic curve.

Answer 2

Multiplying by $\,3\,$ then "completing the square" for the quadratic in $\,x\,$:

$$ \begin{align} & 9x^2+9y^2-12xy+30x-30y+30 \\ = \;\; &9x^2 - 6\, (2y-5)\,x \color{red}{+(2y-5)^2-(2y-5)^2} +9y^2-30y+30 \\ = \;\;&\big(3x-(2y-5)\big)^2 + 5y^2-10y+5 \\ = \;\;&(3 x - 2 y + 5)^2 + 5 (y-1)^2 \end{align} $$

It follows that the only real solution is $\,y=1, x=-1\,$.

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[ EDIT ] $\;$ The final equation can be written as $\,x'^{\,2}+ y'^{\,2}=0\,$ with $\,x' = 3x-2y+5\,$ and $\,y' = \sqrt{5} y-\sqrt{5}\,$, which is a degenerate conic, specifically two intersecting complex lines $\,x' \pm i\, y' = 0\,$ with the single common real point at $\,x' = y' = 0\,$. This is consistent with OP's findings of discriminant $\,\le 0\,$ and center of the conic at the unique real point.

Answer 3

I assume that what you are describing is what you found upon differentiating the conic equation "head-on": $$ \frac{d}{dx} \ [ \ 3x^2 + 3y^2 - 4xy + 10x - 10y + 10 \ ] \ \ = \ \ \frac{d}{dx} \ [ \ 0 \ ] $$ $$ \Rightarrow \ \ 6x \ + \ 6yy' \ - \ 4·[y + xy'] \ + \ 10 \ - \ 10y' \ + \ 10 \ \ = \ \ 0 \ \ \Rightarrow \ \ y' \ \ = \ \ \frac{3x \ - \ 2y \ + \ 5}{2x \ - \ 3y \ + \ 5} \ \ . $$

If we didn't know this is a "degenerate" conic, we could consider the locations of the "horizontal" and "vertical" tangents to the curve. The horizontal tangents are found where $ \ y' \ = \ 0 \ \Rightarrow \ 3x - 2y + 5 \ = \ 0 \ $ intersects the curve and the vertical tangents, where $ \ y' \ $ undefined $ \ \Rightarrow \ 2x - 3y + 5 \ = \ 0 \ $ meets the curve. (An example of how this works for an ordinary conic is shown in my answer here.) We observe, however, that when we insert these relations into the conic equation, we obtain

$$ 3x^2 \ + \ 3·\left(\frac{3x + 5}{2} \right)^2 \ - \ 4x·\left(\frac{3x + 5}{2} \right) \ + \ 10x \ - \ 10·\left(\frac{3x + 5}{2} \right) \ + \ 10 $$ $$ = \ \ \frac{15}{4}·(x + 1)^2 \ \ = \ \ 0 \ \ , $$ $$ 3x^2 \ + \ 3·\left(\frac{2x + 5}{3} \right)^2 \ - \ 4x·\left(\frac{2x + 5}{3} \right) \ + \ 10x \ - \ 10·\left(\frac{2x + 5}{3} \right) \ + \ 10 $$ $$ = \ \ \frac53·(x + 1)^2 \ \ = \ \ 0 \ \ . $$

So both the horizontal and vertical tangents occur at $ \ (-1 \ , \ 1) \ \ (!) \ \ . \ $ Moreover, this is where the two lines intersect. Ordinarily, this intersection would mark the center of the conic section, so there is something decidedly peculiar about this "curve" if this "center" is where vertical and horizontal lines are tangent there as well.

Now that we have the coordinates, we further discover that $ \ (-1 \ , \ 1) \ $ solves the equation, $$ 3·(-1)^2 \ + \ 3·1^2 \ - \ 4·(-1)·1 \ + \ 10·(-1) \ - \ 10·1 \ + \ 10 \ \ = \ \ 3 + 3 + 4 - 10 - 10 + 10 \ \ = \ \ 0 \ \ $$ and that the "slope" given by our derivative function, $ \ y' \ = \ \frac{3·(-1) \ - \ 2·1 \ + \ 5}{2·(-1) \ - \ 3·1 \ + \ 5} \ = \ \frac00 \ \ , $ is indeterminate. So there is only the single point in the real plane, at which the first derivative has no meaningful value.

This is also indicated by the "discriminants" mentioned (see for instance, here concerning degenerate conics). You presumably calculated something along the lines of $ \ \det \mathsf{M} \ = \ \left| \begin{array}{cc} 3 & -4/2 \\ -4/2 & 3 \end{array} \right| \ = \ 5 \ > \ 0 \ \ , $ telling us that the conic is an ellipse. However, the fuller determinant is $$ \ \det \mathsf{Q} \ = \ \left| \begin{array}{ccc} 3 & -4/2 & 10/2 \\ -4/2 & 3 & -10/2 \\ 10/2 & -10/2 & 10 \end{array} \right| \ = \ 0 \ \ , $$ the singularity of this matrix indicating that the ellipse is a single point.

Regarding the remarks about this curve in complex coordinates, we can look at what happens in $ \ \mathbb{C}^2 \ \rightarrow \ \mathbb{R}^4 \ $ by using $ \ x \ = \ a + bi \ \ , \ \ y \ = \ c + di \ $ in the curve equation: $$ 3a^2 \ - \ 3b^2 \ + \ 3c^2 \ - \ 3d^2 \ + \ 10a \ - \ 10c \ - \ 4ac \ + \ 4bd \ + \ 10 \ \ = \ \ 0 \ \ , $$ $$ i \ · \ (6ab \ - \ 4ad \ - \ 4bc \ + \ 6cd \ + \ 10b \ - \ 10d ) \ \ = \ \ 0 \ \ . $$ We would need to take various "slices" through this four-dimensional object described by the curve equation. If we set the imaginary parts of $ \ x \ $ and $ \ y \ $ to zero $ \ [b \ = \ d \ = \ 0 ] \ \ , $ the imaginary equation vanishes and the real equation becomes $ \ 3a^2 \ + \ 3c^2 \ + \ 10a \ - \ 10c \ - \ 4ac \ + \ 10 \ \ = \ \ 0 \ \ , $ which is what we've solved with $ \ a \ = \ -1 \ \ , \ \ c \ = \ 1 \ \ . $ Taking the $ \ a \ = \ c \ = \ 0 \ $ "slice" instead yields $ \ -3b^2 \ - \ 3d^2 \ + \ 4bd \ + \ 10 \ \ = \ \ 0 \ \ , \ \ i \ · \ ( 10b \ - \ 10d ) \ \ = \ \ 0 \ \ , $ which are solved simultaeously by $ \ b \ = \ d \ = \ \pm \sqrt5 \ \ . $ So we have found three points "on" the conic section, $ \ (-1 \ , \ 1) \ $ [the single "real" point), $ \ \ (i\sqrt5 \ , \ i\sqrt5) \ \ , \ \ (-i\sqrt5 \ , \ -i\sqrt5) \ \ . $ Other "slices" reveal other parts of the structure of the figure described by our equation, but this begins to get rather beyond the intent of the orginal problem.