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TANGENTS AND NORMALS

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Slope of Tangent and Normal

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Slope of a tangent

Slope of a tangent to the curve y = f(x) is given by dy / dx




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Slope of a normal

Slope of the normal to the curve y = f(x) is given by -1 / (dy/dx)






Equation of Tangent and Normal

Equation of tangent at (x0, y0)

Equation of the tangent at the point (x0, y0) is given by
                                  (y - y0) = dy / dx (x - x0)
                                              OR
                                  (y - y0) = f’(x) (x - x0)

Equation of normal at (x0, y0)

 
Equation of the normal at the point (x0, y0) is given by
                               (y – y0) = -1 / (dy / dx) (x - x0)
                                               OR
                               (y - y0) = [-1 / f’(x0)] (x - x0)

Condition for perpendicularity and parallelism

•    If the lines are parallel then the slopes are equal
•    If the lines are perpendicular the product of their slopes is -1
•    If the tangent is parallel to x-axis then dy / dx = 0
•    If the tangent is parallel to y-axis then dy / dx = 1 / 0
•    Slope of the line segment joining two points (x1, y1) and (x2, y2) is given by (y2 - y1) / (x2 - x1)
The following examples will help us to understand the concept more thoroughly:
Example 1: Find the point on the curve y = x3 - 11x + 5 at which tangent is y = x - 11
Solution: Slope of the tangent to the curve, dy / dx = 3x2 - 11
Since y = x - 11, dy / dx = 1
3x2 – 11 = 1
x2 = 4
x = 2
When x = 2, y = 23 – 11 x 2 + 5 = 8 – 22 + 5 = -9, so the point is (2, -9)

Example 2: Find the equation of the tangent line to the curve y = x2 - 2x + 7 which is parallel to the line 2x – y + 9 = 0
Solution: Slope of the tangent = 2x - 2
Slope of the line = 2
Since they are parallel, 2x – 2 = 2
x =2
When x = 2, y = 4 – 4 + 7 = 7
So the point is (2, 7)
Equation of the tangent is, y – 7 = 2 (x - 2)
                               y = 2x + 3

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