EQUATION OF TANGENT WHICH IS PARALLEL OR PERPENDICULAR TO THE LINE

Tangent :

The tangent line (or simply tangent) to a plane curve at a given point is the straight line that just touches the curve at that point.

Equation of tangent :

(y-y1)  =  m(x-x1)

Normal :

The normal at a point on the curve is the straight line which is perpendicular to the tangent at that point. The tangent and the normal of a curve at a point are illustrated in the adjoining figure.

Equation of normal :

(y-y1)  =  (-1/m)(x-x1)

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  =  -1/3x ------(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

When x  =  2

y  =  1+23

y  =  9

When x  =  -2

y  =  1+(-2)3

y  =  -7

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)  ---(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

x-1  =  2

x  =  3

y  =  (3+1)/(3-1)

y  =  2

 x - 1  =  -2

x  =  -1

y  =  (-1+1)/(-1-1)

y  =  0

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

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