Find real solution to the equation3x2+3y2−4xy+10x−10y+10=0.What does partial differentiation give for a second degree equation...

m4tx45w

m4tx45w

Answered

2022-01-30

Find real solution to the equation
3x2+3y24xy+10x10y+10=0.
What does partial differentiation give for a second degree equation which doesn't represent a conic?

Answer & Explanation

ocretz56

ocretz56

Expert

2022-01-31Added 16 answers

Step 1
A conic curve, is the intersection of the 3D cone
p1z2=p2x2+p3y2
with the 3D plane
p4x+p5y+p6=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
z2=x2+y2 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

Gordon Stephens

Expert

2022-02-01Added 10 answers

Step 1
Multiplying by 3 then "completing the square" for the quadratic in x:
9x2+9y212xy+30x30y+30
=9x26(2y5)x+(2y5)2(2y5)2+9y230y+30
=(3x(2y5))2+5y210y+5
=(3x2y+5)2+5(y1)2
It follows that the only real solution is y=1, x=1
Step 2
The final equation can be written as
x2+y2=0 with x=3x2y+5 and y=5y5,
which is a degenerate conic, specifically two intersecting complex lines x±iy=0 with the single common real point at x=y=0. This is consistent with OP's findings of discriminant 0 and center of the conic at the unique real point.

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