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).
For the curve y = x²+x, at what point does the normal line at (0, 0) intersect the tangent line at (1, 2)?
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