I don't know to find parabola mx^2+2mx+x+m-1. all i know is the extreme of parabola is on the line 9x-3y-3.

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The extreme of the parabola is determined by finding the first derivative and equating it to 0. We then solve for x.

y = mx^2+2mx+x+m-1

y' = 2mx + 2m + 1 = 0

=> 2mx = -2m - 1

=> x = (-2m - 1) / 2m

For x = (-2m - 1) / 2m

y = m*(-2m - 1)^2/ 4m^2 + (2m +1)*(-2m - 1) / 2m + m-1

Note : 9x - 3y - 3 is not the equation of a line, I have assumed that you meant 9x - 3y = 3 or 3x - y = 1

Now this point lies on the line 3x - y = 1

=> 3*(-2m - 1) / 2m  - m*(-2m - 1)^2/ 4m^2 - (2m +1)*(-2m - 1) / 2m - m + 1 = 1

=> 3*(-2m - 1) / 2m  - m*(-2m - 1)^2/ 4m^2 - (2m +1)*(-2m - 1) / 2m - m = 0

=> -3*(2m + 1) / 2m  + (2m +1)^2 / 4m - m = 0

=> -6*(2m + 1) / 4m  + (2m +1)^2 / 4m - m*4m / 4m = 0

=> -12m - 6 + 4m^2 + 1 + 4m - 4m^2 = 0

=> -8m - 5 = 0

=> m = -5/8

Therefore the parabola is mx^2+2mx+x+m-1

=> y = (-5/8)x^2 - (5/4)x + x - (13/8)

The required parabola is y = (-5/8)x^2 - (5/4)x + x - (13/8)

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