# EQUATION OF TANGENT AND NORMAL OF CONIC SECTIONS WITH GIVE PARAMETER

To find equation of the tangent line, we should be aware of the parametric form

Circle

## Parametric equations

x = a cos θ, y = a sin θ

Parabola

x = at2, y = 2at

Ellipse

x = a cos θ, y = a sin θ

Hyperbola

x = a sec θ, y = b tan θ

In the given conic  function, we have to do the following changes.

x2 = xx1, y2 = yy1

x = (x + x1)/2 and y = (y + y1)/2

Applying the point that we have derived from parametric equation in this equation, we will get equation of the tangent line.

ax + by = c

To find the normal, we have to interchange the coefficients of x and y. That is,

bx - ay = k

Applying x and y, we can solve for k.

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

 Finding at2a = 2 and t = 1/2= 2(1/2)2= 2(1/4)= 1/2 Finding 2ata = 2 and t = 1/2= 2(2)(1/2)= 4/2= 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.

 a2 = 32 a = √32a = √(16 × 2)a = 4√2x = a cos  θx = 4√2 cos (π/4)= 4√2 × (1/√2)x = 4 b2 = 8b = √8b = √(4 × 2)b = 2√2y = b sin  θy = 2√2 sin (π/4)= 2√2 × (1/√2)y = 2

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 :

 a2 = 9a = 3x = a sec  θx = 3 sec (π/6)= 3 (2/√3)x = 2√3 b2 = 12b= 2√3y = b tan  θy = 2√3 tan (π/6)= 2√3 (1/√3)y = 2

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