at what point does the line normal to y = 3 +2x + 3x^2 at (1, 8) intersect the parabola a second time?

y = 3 +2x + 3x^2   (1)     ..... at (1, 8)

The slope of the line at any point on the parabola is given by :

y '  =  2 + 6x

So....at (1,8)  the slope is

y'(1)  = 2 + 6(1)  = 8

And the line normal to this will have the slope  of -1/8   ....so......the equation of this line is

y = (-1/8)(x - 1) + 8

y = (-1/8)x + 65/8      (2)

So.....setting (1)  = (2)   we have

3 + 2x + 3x^2  = (-1/8)x + 65/8

24 + 16x + 24x^2  = -x + 65

24x^2 + 17x - 41 = 0       factor

(24x  + 41) (x -1) = 0

We already know that setting the second factor to 0  produces a correct answer

Setting the first factor to 0 produces

x = -41/24

And putting this into

y = (-1/8)x  + 65/8   produces  1601/192

So....the second point of intersection  is (-41/24, 1601/192)

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