conic sections

Asked by gazalbhatnagar | 20th Feb, 2009, 07:18: PM

Expert Answer:

let eqn is x2 + y2 +2gx + 2fy + c =0   it passes through (4,1) so we have  8g + 2f + c + 17 = 0 .....(1)

passes through (6,5)  so we have  12g + 10 f + c + 61 = 0.........(2)

center (-g , -f ) on the line 4x + y = 16  ===>  4g + f = -16..........(3)

solving we get  g = -3 , f = -4 , c  = 15  and eqn as  x2 + y2 - 6x - 8 y +15 = 0

Answered by  | 20th Feb, 2009, 07:34: PM

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