m4tx45w

2022-01-30

Find real solution to the equation
$3{x}^{2}+3{y}^{2}-4xy+10x-10y+10=0.$
What does partial differentiation give for a second degree equation which doesn't represent a conic?

ocretz56

Expert

Step 1
A conic curve, is the intersection of the 3D cone
${p}_{1}{z}^{2}={p}_{2}{x}^{2}+{p}_{3}{y}^{2}$
with the 3D plane
${p}_{4}x+{p}_{5}y+{p}_{6}=0$
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.

Gordon Stephens

Expert

Step 1
Multiplying by 3 then "completing the square" for the quadratic in x:
$9{x}^{2}+9{y}^{2}-12xy+30x-30y+30$
$=9{x}^{2}-6\left(2y-5\right)x+{\left(2y-5\right)}^{2}-{\left(2y-5\right)}^{2}+9{y}^{2}-30y+30$
$={\left(3x-\left(2y-5\right)\right)}^{2}+5{y}^{2}-10y+5$
$={\left(3x-2y+5\right)}^{2}+5{\left(y-1\right)}^{2}$
It follows that the only real solution is
Step 2
The final equation can be written as
${x}^{\prime 2}+{y}^{\prime 2}=0$ with ${x}^{\prime }=3x-2y+5$ and ${y}^{\prime }=\sqrt{5}y-\sqrt{5}$,
which is a degenerate conic, specifically two intersecting complex lines ${x}^{\prime }±i{y}^{\prime }=0$ with the single common real point at ${x}^{\prime }={y}^{\prime }=0$. This is consistent with OP's findings of discriminant $\le 0$ and center of the conic at the unique real point.

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