# 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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### 2 Answers

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)**

To determine the parabola, we'll have to find out the unknown m.

The extreme point of the parabola is the vertex of the parabola.

If the vertex of the parabola belongs to the line 9x-3y-3 = 0, then the coordinates of the vertex verify the equation of the line.

Let's note the vertex as V(xV, yV)

xV = -b/2a

yV = -delta/4a

We'll substitute xV and yV in the equation of the line:

9xV - 3yV - 3 = 0

We'll divide by 3:

3xV - yV - 1 = 0

We'll put the equation in the slope intercept form:

yV = 3xV - 1

-delta/4a = -3b/2a - 1

delta = b^2 - 4ac

We'll identify a,b,c:

a = m

b = 2m+1

c = m-1

delta = (2m+1)^2 - 4m(m-1)

We'll expand the square and we'll remove the brackets:

4m^2 + 4m + 1 - 4m^2 + 4m

We'll eliminate like terms:

delta = 8m + 1

-delta/4a = -3b/2a - 1

-(8m+1)/4m = -3(2m+1)/2m - 1

-8m-1 = -12m - 6 - 4m

8m + 5 = 0

We'll add -5 both sides:

8m = -5

We'll divide by 8:

m = -5/8

**The parabola is: -5x^2/8 - x/4 - 13/8**