# EQUATION OF TANGENT LINE DRAWN TO THE CURVE

Problem 1 :

Find the equation of the two tangent can be drawn from (5, 2) to the ellipse 2x2 + 7y2 = 14

Solution :

2x2 + 7y2 = 14

Equation of tangent line from the point (x, y) will be

y = mx ±√a2m2+b2

Equation of tangent :

While applying the values of m, we get

Problem 2 :

Find the equations of tangents to the hyperbola

which are parallel to 10x - 3y + 9 = 0

Solution :

a2 = 16, b2 = 64

From the given line, we find the value of m.

3y = 10x + 9

y = (10/3)x + 3

m = 10/3

Problem 3 :

Show that the line x - y + 4 = 0 is a tangent to the ellipse

x2 + 3y2 = 12

Also find the coordinate of the point of contact.

Solution :

If the line y = m x + b is a tangent to the given ellipse it has to satisfy the following condition.

c2 = a2m2 + b2

x - y + 4 = 0

y = x + 4

m = 1 and c = 4

42 = 12(1)2 + 4

16 = 12 + 4

16 = 16

So, the given line is tangent to the given ellipse.

Problem 4 :

Find the equation of the tangent to the parabola y2 = 16x perpendicular to 2x + 2y + 3 = 0

Solution :

2x + 2y + 3 = 0

2y = -2x - 3

y = -x - (3/2)

m = -1

Slope of the tangent line drawn to the curve = 1

y2 = 16x can be compared with y2 = 4ax

4a = 16 and a = 4

Equation of tangent line :

y = mx +(a/m)

y = 1x + (4/1)

y = x + 4

Problem 5 :

Find the equation of the tangent at t = 2 to the parabola y2 = 8x

Solution :

y2 = 8x

Point of the tangent at t = 2 for a parabola will be (at2, 2at)

y2 = 8x

y2 = 4ax

4a = 8 and a = 2

 Finding at2a = 2 and t = 2= 2(2)2= 8 Finding 2ata = 2 and t = 2= 2(2)(2)= 8

Equation of the tangent to the parabola :

y = mx + a/m ---(1)

Applying (8, 8) to find m.

8 = m(8) + 2/m

8 - 8m = 2/m

8m - 8m2 = 2

4m - 4m2 = 1

4m2 - 4m + 1 = 0

(2m - 1) (2m - 1) = 0

m = 1/2 and m = 1/2

Applying m = 1/2 in (1), we get

y = (1/2)x + 2/(1/2)

y = x/2 + 4

2y = x + 8

x - 2y + 8 = 0

Problem 6 :

Find the equations of tangent and normal to hyperbola 12x2 - 9y2 = 108 at θ = π/3

Solution :

Equation of tangent for the parabola :

12xx1 - 9yy1 = 108

At (6, 6)

12x(6) - 9y(6) = 108

72x - 54y = 108

Dividing by 6, we get

12x - 9y = 18

Dividing by 3, we get

4x - 3y = 6

Equation of normal :

3x + 4y = k

At (6, 6)

3(6) + 4(6) = k

k = 18 + 24

k = 42

3x + 4y = 42

So, equation of tangent is 4x - 3y = 6 and normal is 3x + 4y = 42.

Problem 7 :

Prove that the point of intersection of the tangents at t1 and t2 on the parabola y2 = 4ax is (at1 t2, a(t1 + t2))

Solution :

Equation of tangent at the point t1 and t2 :

Equation of tangent for the parabola

yt = x + at2

Equation of tangent at the point t1 and t2 are yt1 = x + at12 and  yt2 = x + at22

yt1 = x + at1---------(1)

yt2 = x + at2---------(2)

(1) - (2)

yt1 - yt= x + at1- (x + at22)

y(t1 - t2) = at1- at22

y(t1 - t2) = a(t1- t22)

y(t1 - t2) = a(t+ t2) (t1 - t2)

y = a(t+ t2)

Applying the value of y in (1)

a(t+ t2)t1 = x + at1

a t2t1 = x

x = a t1t2

Point of intersection is (a t1t2, a(t+ t2)).

Problem 8 :

If the normal at the point t1 on the parabola y2 = 4ax meets the parabola again at the point t2 , then prove that

t2 = -(t1 + 2/t1)

Solution :

Equation of normal to the parabola is

y + xt = 2at + at3

It passes through t1.

y + x t1 = 2a t1 + a t1-------(1)

Any point on the parabola will be in the form (at2, 2at)

It passes through t2.

(at2, 2at) ==> (at22, 2at2)

2at2 + at22 t1 = 2a t1 + a t1------(2)

2at2 - 2a t1 = a t13at22 t1

2a(t2 - t1) = at1  (t12- t22)

2a(t2 - t1) = -at1  (t22- t12)

2a(t2 - t1) = -at1 (t1- t2)(t1+ t2)

2a= -a t1 (t1+ t2)

2 = - t1 (t1+ t2)

(t1+ t2) = -2/ t1

t= -2/t1 - t1

t= -(2/t1 + t1)

## Recent Articles 1. ### Solving Direct and Inverse Proportion Word Problems Worksheet

Sep 22, 23 08:41 AM

Solving Direct and Inverse Proportion Word Problems Worksheet

2. ### Round the Decimals to the Nearest Indicated Place Value

Sep 22, 23 06:13 AM

Round the Decimals to the Nearest Indicated Place Value