The normal to the curve y=2x^2-x+3 at the point P(1,4) meets the curve again at the point Q.

Question

What is the x-coordinate of q?

Answer 1

The curve y = 2x² - x + 3

Differentiating  on each side with respect of x.

y' = 4x - 1

Substitute the value x = 1 in above equation.

y' = 4(1) - 1

y' = 4 - 1

y' = 3

This is the slope of tangent line to the curve at (1, 4).

Tangent line and normal line are perpendicular.

Perpendicular lines slopes are negative reciprocal to each other.

So - 1/3 slope of normal line to the curve at (1, 4).

To find the normal line equation, substitute the values of m = (-1/3) and (x, y) = (1, 4).  in the slope intercept form of an equation y = mx + b.

4 = (-1/3)1 + b

b = 4 + (1/3)

b = 13/3

Substitute m = -1/3 and b = 13/3 in y = mx + b.

y = (-1/3)x + (13/3)

Normal line equation is y = (- x/3) + (13/3).

To find intersect point of curve and normal, equate the two equations.

(- x/3) + (13/3) = 2x² - x + 3

 ( - x + 13)/3 = 2x² - x + 3

- x + 13 = 6x² - 3x + 9

6x² - 3x + 9 + x - 13 = 0

6x² - 2x - 4 = 0

Divide each side b y 2.

3x² - x - 2 = 0

3x² - 3x + 2x - 2 = 0

3x(x - 1) + 2(x - 1) = 0

(x - 1)(3x + 2) = 0

x = 1 and x = - 2/3

The x coordinates are 1 and -2/3.

y = (- 1/3)(1) + (13/3) =  4

 y = (- 1/3)(-2/3) + (13/3) = 41/9.

The points are (1, 4) and (-2/3 , 41/9)

P is (1, 4).

Therefore, the x coordinate of Q = -2/3.