FIND EQUATION OF TANGENT TO PARABOLA

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, yis -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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