SOLUTION: Can anybody give me some hint about this question ? The conic section is given x^2+y^2-2xy+10x-2y+9=0 *find the canonical equation and the particular type of the conic secti

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can anybody give me some hint about this question ? The conic section is given x^2+y^2-2xy+10x-2y+9=0 *find the canonical equation and the particular type of the conic secti      Log On


   

Question 1085163: Can anybody give me some hint about this question ?
The conic section is given x^2+y^2-2xy+10x-2y+9=0
*find the canonical equation and the particular type of the conic section k ?
Found 2 solutions by ikleyn, Alan3354:
Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
HINT: x%5E2+-+2xy+%2B+y%5E2 = %28x-y%29%5E2. 

Therefore,

x%5E2%2By%5E2-2xy%2B10x-2y%2B9 = 0  is equivalent to

%28x-y%29%5E2+%2B+10x+-+2y = -9.   (1)


Now introduce new variables

u = x - y,   (2)
v = x + y.   (3)


From (2) and (3) you have

x = %28u+%2B+v%29%2F2,    (4)
y = %28-u+%2B+v%29%2F2.   (5)


Now substitute these things into (1). You will have

u%5E2+%2B+10%28%28u%2Bv%29%2F2%29+-+2%28%28v-u%29%2F2%29 = -9,    or, equivalently,

u%5E2+%2B+5%28u%2Bv%29+-+%28v-u%29 = -9,    or, equivalently,

u%5E2+%2B+6u+%2B+4v = -9, 

%28u%2B3%29%5E2+%2B+4v = -9+%2B+9,

%28u%2B3%29%5E2+%2B+4v = 0,

%28u%2B3%29%5E2 = -4v.


Which conic section is this ??


Correct !  It is a PARABOLA, in coordinates (u,v).

Solved.



Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
You have to rotate the axes wrt the conic.
Do you know how to do that?