For the curve y = x²+x, at what point does the normal line at (0, 0) intersect the tangent line at (1, 2)?
To find the point of intersection of the normal line at (0,0) and the tangent line at (1,2) for the curve y = x² + x, we need to determine the equations of both lines and then find their point of intersection.
First, let's find the tangent line at (1,2):
The slope of the tangent line is the derivative of the curve at point (1,2). Taking the first derivative of y = x² + x:
dy/dx = 2x + 1
Now, substitute x = 1 into the derivative equation to find the slope at (1,2):
dy/dx = 2(1) + 1 = 3
So, the slope of the tangent line is 3.
Using the point-slope form of a line, we can write the equation of the tangent line at (1,2):
y - 2 = 3(x - 1)
Simplifying the equation, we get:
y - 2 = 3x - 3
y = 3x - 1
Now, let's find the normal line at (0,0):
The slope of the normal line is the negative reciprocal of the tangent line, which is -1/3.
Using the point-slope form of a line, we can write the equation of the normal line at (0,0):
y - 0 = -1/3(x - 0)
Simplifying the equation, we get:
y = -1/3x
Now, to find the point of intersection, we can set the y values of both lines equal to each other:
3x - 1 = -1/3x
Multiplying both sides of the equation by 3 to eliminate the fraction:
9x - 3 = -x
Simplifying the equation, we get:
10x = 3
Dividing both sides of the equation by 10:
x = 3/10
Substituting this x value back into one of the lines' equations, let's find the y value:
y = -1/3(3/10)
y = -1/10
Therefore, the point of intersection of the normal line at (0,0) and the tangent line at (1,2) for the curve y = x² + x is (3/10, -1/10).