EQUATION OF TANGENT WHICH IS PARALLEL OR PERPENDICULAR TO THE LINE

Problem 1 :

Find the equations of the tangents to the curve

y  =  1+x³

for which the tangent is orthogonal with the line

x+12y  =  12

Solution :

y  =  1+x³

Slope of tangent drawn at the point (x, y).

dy/dx  =  3x²

Slope of normal  =  -1/3x²  ------(1)

The tangent line is perpendicular to the given line

x+12y  =  12

y  =  -x/12 + 1

Slope (m)  =  -1/12 ------(2)

(1)  =  (2)

-1/3x²  =  -1/12

3x²  =  12

x²  =  4

x  =  2, -2

y  =  1+x³

So, the required points are (2, 9) and (-2, -7)

Equation of tangent :

Slope of tangent at the point (2, 9) is 

m  =  3(2)²

m  =  12

Equation of tangent at the point (2, 9) :

(y-9)  =  12(x-2)

y-9  =  12x-24

12x-y-24+9  =  0

12x-y-15  =  0

Equation of tangent at the point (-2, -7) :

(y+7)  =  12(x+2)

y+7  =  12x+24

12x-y+24-7  =  0

12x-y+17  =  0

Problem 2 :

Find the equations of the tangents to the curve

y  =  (x+1)/(x-1)

which are parallel to the line x+2y  =  6.

Solution :

y  =  (x+1)/(x-1)

Differentiating with respect to x.

Using quotient rule,

u  =  x+1 and v  =  x-1

u'  =  1 and v'  =  1

dy/dx  =  (x-1)(1)-(x+1)(1) / (x-1)²

=  (x-1-x-1) / (x-1)²

=  -2/(x-1)²    ---(1)

Slope of the given line :

x+2y  =  6

y  =  -x/2+3

m  =  -1/2---(2)

Since the tangent drawn to the curve at the point (x, y) is parallel to the given line

(1)  =  (2)

-2/(x-1)²  =  -1/2

4  =  (x-1)²

(x-1)  =  ±2

Equation of the tangent passes through (3, 2) :

y-2  =  (-1/2)(x-3)

2(y-2)  =  -1(x-3)

2y-4  =  -x+3

x+2y-4-3  =  0

x+2y-7  =  0

Equation of the tangent passes through (-1, 0) :

y-0  =  (-1/2)(x+1)

2y  =  -1(x+1)

2y  =  -x-1

x+2y+1  =  0

Problem 3 :

Find the equation of tangent and normal to the curve given by

x  =  7 cos t and y  =  2 sin t for all t

at any point on the curve.

Solution :

Slope of tangent :

dx/dt  =  -7 sin t and dy/dt  =  2 cos t

dy/dx  =  2 cost/(-7 sin t)

dy/dx  =  (-2/7) (cost/sin t)

Equation of tangent :

 (y-2sin t)  =  (-2/7) (cost/sin t)(x-7cost)

7sint (y-2sin t)  =  (-2cost)(x-7cost)

(2cost) x + (7sint)y  =  14cos²t + 14sin²t

(2cost) x + (7sint)y  =  14(cos²t + sin²t)

(2cost) x + (7sint)y  =  14(1)

(2cost) x + (7sint)y  =  14

Equation of normal :

 (y-2sin t)  =  (7/2) (sin t/cos t)(x-7cost)

2cost (y-2sin t)  =  (7sint)(x-7cost)

(2cost)y-4sint cost  =  (7sint)x - 49sint cost

(7sint)x - (2cost)y - 49sint cost + 4sintcost  =  0

(7sint)x - (2cost)y - 45sint cost  =  0

Problem 4 :

Find the equation of the tangent to the curve

y = x² – 2x + 7

which is perpendicular to the line 5y - 15x = 13

Solution :

y = x² – 2x + 7

Differentiate with respect to x, we get

dy/dx = 2x - 2(1) + 0

dy/dx = 2x - 2

Slope of the given line 5y - 15x = 13

5y = 15x + 13

y = 3x + (13/5)

Slope = 3

Slope of the perpendicular line = -1/3

2x - 2 = -1/3

6x - 6 = -1

6x = -1 + 6

6x = 5

x = 5/6

When x = 5/6, we get y  = (5/6)² – 2(5/6) + 7

= 25/36 - 5/3 + 7

= (25 - 60 + 252)/36

= 217/36

Equation of tangent :

y - 217/36 = -1/3(x - 5/6)

y - 217/36 = -x/3 + 5/18

y = -x/3 + 5/18 + 217/36

y = (-12x + 10 + 217)/36

36y = -12x + 227

12x + 36y = 227

Problem 5 :

Find the equation of the normal to the curve y = x³ +2x +6 which are parallel to the line x + 14y + 4 = 0

Solution :

Normal will be parallel to line x + 14y + 4 = 0

14y = - x - 4

y = (-1/14)x - 2/7

Slope = -1/14

y = x³ +2x +6

dy/dx = 3x² + 2(1) + 0

slope of tangent = 3x² + 2

Slope of normal = -1/(3x² + 2)

-1/(3x² + 2) = -1/14

3x² + 2 = 14

3x² = 12

x² = 4

x = 2 and x = -2

When x = 2, y = 2³ +2(2) +6

y = 18

When x = -2, y = (-2)³ +2(-2) +6

y = -12 + 6

y = -6

So, the required points are (2, 18) and (-2, -6)

Equation of normal :

y - 8 = -1/14 (x - 2)

14y - 112 = -x + 2

x + 14y - 112 - 2 = 0

x + 14y - 114 = 0

Equation of normal :

y + 6 = -1/14 (x + 2)

14y + 84 = -x - 2

x + 14y + 84 + 2 = 0

x + 14y + 86 = 0

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