Analytic geometry

Question

Determine the type, draw graph and find the equation of central quadratic that symetric to the xyz plane and passing through (1,-2,1), (-1,0,3), (2,0,0) points?

Answer 1

symmetric means

(1,-2,1) ---> (-1,2,1)

(-1,0,3)---> (1,0,3)

(2,0,0)-->(-2,0,0)

y = f(x,y) = Ax^2 + Bt^2 + Cx + kT + Lambda

where A,B,C,k, and L=Lambda are fixed number constants

the original system is:

1 = A + 4B + C - 2k + L

3 = A - C +L

0 = 4A + 2C + L

1 = A +4B - C + 2K + L

3 = A + C + L

the 6th point (-2,0,0) results in the same equation 0= 4a + 2c + L

step 1:

Eq 5 - eq 2

0 = 2c

c=0

the system becomes

1 = a + 4b - 2k + L

3 = a +L

0 = 4a +L

1 = a + 4b + 2k + L

step 2:

eq4 - eq 1

0 = 4k

k=0

the system becomes

1 = a + 4b + L

3 = a + L

0 = 4a + L

step 3:

eq 1 - eq 2

-2 = 4b

b = -1/2

the system becomes:

3 = a + L

0= 4a + L

step 4:

eq 2 - eq 1

-3 = 3a

a = -1

then 0 = 4(-1) + L

0 = -4 + L

L = 4

the function is y = f(x,t) = -x^2 - (1/2)t^2 + 4