A tangent to a parabola is a straight line which intersects (touches) the parabola exactly at one point.
Example 1 :
Determine the equation of the tangent to the curve defined by
f(x) = x3+2x2-7x+1
at x = 2.
Solution :
f(x) = x3+2x2-7x+1
When x = 2
y = 8+8-14+1
y = 17-14
y = 3
Slope of tangent :
f'(x) = 3x2+4x-7
f'(2) = 3(4)+4(2)-7
= 12+8-7
f'(2) = 13
Equation of tangent :
(y-3) = 13(x-3)
y-3 = 13x-39
y = 13x-36
Example 2 :
Find an equation of the tangent line drawn to the graph of
y = x 2-9x+7
with slope -3.
Solution :
y = x 2-9x+7
Let (x, y) be the point where we draw the tangent line on the curve.
Slope of tangent at point (x, y) :
dy/dx = 2x-9
Slope of the required tangent (x, y) is -3.
2x-9 = -3
2x = 6
x = 3
By applying the value of x in y = x 2-9x+7
y = 9-27+7
y = -11
Equation of the tangent line :
y-y1 = m(x-x1)
y+11 = -3(x-3)
y+11 = -3x+9
3x+y+11-9 = 0
3x+y+2 = 0
Equation of the tangent line is 3x+y+2 = 0.
Example 3 :
Find a point on the curve
y = x2-2x-3
at which the tangent is parallel to the x axis.
Solution :
y = x2-2x-3
If the tangent line is parallel to x-axis, then slope of the line at that point is 0.
Slope of the tangent line :
dy/dx = 2x-2
2x-2 = 0
2x = 2
x = 1
By applying the value x = 1 in y = x2-2x-3, we get
y = 1-2-3
y = -4
So, the required point is (1, -4).
Example 4 :
Determine the point(s) on the curve
f(x) = (2x-1)2
where the tangent is
(i) parallel to the line y = 4x-2
(ii) Perpendicular to the line 2y+x-4 = 0
Solution :
f(x) = (2x-1)2
f'(x) = 2(2x-1)(2-0)
f'(x) = 4(2x-1) ---(1)
If two lines are parallel, then slopes will be equal.
(i)
y = 4x - 2 is the line which is parallel to the tangent line.
Slope of y = 4x - 2 :
m = 4 ---(2)
Slope of the tangent line at the point (x, y) is
m = 4(2x-1)
(1) = (2)
4(2x-1) = 4
2x-1 = 1
2x = 2
x = 1
By applying the value of x in y = (2x-1)2, we get
y = 1
So, the required point is (1, 1).
(ii) If two lines are perpendicular, then the product of their slopes will be equal to -1.
Slope of the perpendicular line 2y+x-4 = 0 is m = -1/2.
Slope of the tangent line = 2 ---(3)
4(2x-1) = 2
2(2x-1) = 1
4x - 2 = 1
4x = 3
x = 3/4
By applying the value of x in f(x) = (2x-1)2, we get
f(3/4) = (3/2 - 1)2
f(3/4) = (1/2)2
f(3/4) = 1/4
So, the required point is (3/4, 1/4).
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