Linear Algebra Question: find all conic equations that match the given points I

Question

Linear Algebra Question: find all conic equations that match the given points

I am trying to find all conic equations that pass through the points:

(0,0), (2, 0), (0, 2), (2, 2), (1, 3), (4, 1).

The form of the conic equation we should use is:

f(x, y) = c1 + c2x + c3y + c4(x2) + c5(xy) + c6(y2)

I made a matrix to solve this problem, but after a long rref process (we have to do this by hand) I found that the solution set was inconsistent. Am I correct or did I make a mistake somewhere? I just want to be sure because the prompt does not give any instructions on what to do if the given points do not match any conic equations.

Thanks in advance for your help!

Explanation / Answer

f(x, y) = c1 + c2x + c3y + c4(x2) + c5(xy) + c6(y2)

so we have the points (0,0), (2, 0), (0, 2), (2, 2), (1, 3), (4, 1).

Hence c1 =0 when we put (0,0) and euqate it to zero

Again similarly substitute one by one and equate them to zero

we get 2*c2 +4*c4 = 0

=> c2 = -2c4...................(a)

again for (0,2) we get

2*c3 +4c6 = 0

c3 = -2c6.......................(b)

For (2,2)

we get 2c2 + 2c3 + 4(c4 +c5+ c6) =0

we get from (a) ,(b) c5 =0 after substitution

again for (1,3) we get

c2 +3*c3 +c4 +c6(9) =0

Again we have from (b) and (a)

-2c4 -6c6 + c4+9c6 =0

=> c4 = 3c6..................(c)

Finally for (4,1) we get

4c2 + c3 +16*c4 + c6 =0

from (a) , (b) ,(c) we get

-8c4 -2c6+16c4 +c6 =0

=> 8c4 = c6

we also got c4 = 3c6 so we get c4=c6=0

Hence c1=c2=c3=c4=c5=c6=0

So no curve passes through all the points

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