EQUATION OF TANGENT AND NORMAL OF CONIC SECTIONS WITH GIVE PARAMETER

Find the equations of the tangent and normal

Problem 1 :

To the parabola y² = 8x at t = 1/2

Solution :

Given, equation of the parabola y² = 8x.

Comparing this equation with y² = 4ax

Point of the tangent at t = 1/2 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 (1/2, 2) to find m.

2 = m(1/2) + 2/m

 2 =  m/2 + 2/m

2 = (m² + 4)/2m

4m = m² + 4

m² - 4m + 4 = 0

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

m = 2 and m = 2

Applying m = 2 in (1), we get

y = 2x + 2/2

y = 2x  + 1

2x - y + 1 = 0

Equation of normal :

2y = x  +  k

At(1, 1)

2(1) = 1 + k

2 = 1 + k

k = 1

2y  = x + 1

x - 2y + 1 = 0

So, equation of tangent is 2x - y + 1 = 0 and normal is x - 2y + 1 = 0.

Problem 2 :

To the ellipse x² + 4y² = 32 at θ = π/4

Solution :

x² + 4y² = 32

Dividing 32 on each sides.

Equation of the tangent :

y - y₁ = m(x - x₁)

y - 2 = -1/2(x - 4)

y - 2 = -x/2 + 4/2

y = -x/2 + 4/2 + 2

y = -x/2 + 4/2 + 4/2

y = (-x + 8)/2

2y = -x + 8

2y + x = 8

x + 2y - 8 = 0

Equation of normal :

2x + y = k

At (4, 2)

2(4) + 2 = k

8 + 2 = k

10 = k

2x + y - 10 = 0

So, equation of tangent is x + 2y - 8 = 0 and normal is 2x - y - 10 = 0.

Problem 3 :

Solution :