EQUATION OF TANGENT LINE WITH PARAMETRIC EQUATIONS

Problem 1 :

Find the equation of tangent and normal to the curve given by x = 7 cos t and y = 2 sin t at any point on the curve.

Solution :

x = 7 cos t and y = 2 sin t

To find equation of tangent, we have to find the slope of the tangent line by finding the derivative of the given curves.

dx/dt = -7 sin t

dy/dt = 2 cos t

dy/dx = 2 cos t / (-7 sin t)

Slope = (-2/7) (cos t / sin t)

Equation of the tangent line :

(y - y₁) = m(x - x₁)

(y - 2 sin t) = (-2/7) (cos t / sin t)(x  - 7 cos t)

sin t(y - 2 sin t) = -2 cos t(x - 7 cos t)

7y sin t - 14 sin² t = -2 x cos t + 14 cos² t

2x cos t + 7y sin t = 14 cos² t + 14 sin² t

2x cos t + 7y sin t = 14 (cos² t + sin² t)

2x cos t + 7y sin t = 14 (1)

2x cos t + 7y sin t = 14

Equation of the tangent line :

(y - y₁) = -1/m(x - x₁)

(y - 2 sin t) = (7/2) (sin t / cost t)(x  - 7 cos t)

2 cos t (y - 2 sin t) = 7 sin t (x - 7 cost)

2y cos t - 4 sin t cos t = 7x sin t - 49 sin t cos t

7x sin t - 2y cos t - 49 sin t cos t + 4 sin t cos t = 0

7x sin t - 2y cos t - 45 sin t cos t = 0

Problem 2 :

x = cos t, y = 2 sin² t at t = π/3

Solution :

Applying the value of t in the given function, we get the value of x and y.

dx/dt = -sin t, dy/dt = 4 sin t cost t

dy/dx = (dy/dt) / (dx/dt)

= (4 sin t cost t) / (-sin t)

dy/dx = -4 cos t 

Slope at t = π/3

dy/dx = -4 cos (π/3)

= - 4(1/2)

Slope of tangent = -2

Slope of normal = 2

Equation of tangent :

(y - y₁) = m(x - x₁)

x = 1/2 and y = 3/2

Slope = -2

(y - (3/2)) = -2 (x - (1/2))

2y - 3 = -2(2x - 1)

2y - 3 = -4x + 2

4x + 2y -3 - 2 = 0

4x + 2y -5 = 0

Equation of normal :

(y - y₁) = (-1/m)(x - x₁)

x = 1/2 and y = 3/2

Slope = 1/2

(y - (3/2)) = 1/2(x - (1/2))

2y - 3 = 1/2(x - 1)

4y - 6 = x - 1

x - 4y - 1 + 6 = 0

x - 4y + 5 = 0