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

Guest Feb 8, 2016

#2**+5 **

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)

Here's a graph : https://www.desmos.com/calculator/sjnizqdyzq

CPhill Feb 8, 2016

#1**0 **

Why don't try this fine graphing calculator and see if it will solve your problem:

https://www.desmos.com/calculator

Guest Feb 8, 2016

#2**+5 **

Best Answer

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)

Here's a graph : https://www.desmos.com/calculator/sjnizqdyzq

CPhill Feb 8, 2016