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

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 + at₁²  ---------(1)

yt₂ = x + at₂² ---------(2)

(1) - (2)

yt₁ - yt₂ = x + at₁² - (x + at₂²)

y(t₁ - t₂) = at₁² - at₂²

y(t₁ - t₂) = a(t₁² - t₂²)

y(t₁ - t₂) = a(t₁ + t₂) (t₁ - t₂)

y = a(t₁ + t₂)

Applying the value of y in (1)

a(t₁ + t₂)t₁ = x + at₁² 

a t₂t₁ = x

x = a t₁t₂

Point of intersection is (a t₁t_2, a(t₁ + 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 t₁³  -------(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 t₁³ ------(2)

2at₂ - 2a t₁ = a t₁³- at₂² 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

t₂ = -2/t₁ - t₁

t₂ = -(2/t₁ + t₁)