Question

Find the equations of tangents and normals to the following curve at the indicated points on them:

x = sin θ and y = cos 2θ at θ = `pi/(6)`

Answer 1

When `theta = pi/(6), x = sin  pi/(6) and y = cos  pi/(3)`

∴ `x = (1)/(2) and y = (1)/(2)`

Hence, the point at which we want to find the equations of tangent and normal is `(1/2, 1/2)`.

Now, x = sinθ, y = cos 2θ

Differentiating x and y w.r.t. θ, we get

`dx/(dθ) = d/(dθ)(sinθ) = cosθ`

and

`dy/(dθ) = d/(dθ)(cos2θ)`

= `-sin2θ.d/(dθ)(2θ)`

= – sin 2θ × 2

= – 2 sin 2θ

∴ `dy/dx = ((dy/(dθ)))/((dx/(dθ))`

= `(-2sin2θ)/(cosθ)`

∴ `(dy/dx)_(""at"  θ  =  pi/6) = (-2sin  pi/3)/(cos  pi/6)`

= `(-2(sqrt(3)/2))/((sqrt(3)/2)`

= – 2

= Slope of the tangent at θ = `pi/(6)`

∴ The equation of the tangent at θ = `pi/(6) "i.e. at"(1/2, 1/2)` is

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

∴ `y - (1)/(2) = -2x + 1`

∴ 2y – 1 = – 4x + 2

∴ 4x + 2y – 3 = 0

The slope of normal at θ = `pi/(6)`

= `-(1)/((dy/dx)_("at" theta  =  pi/6)`

= `(-1)/(-2)`

= `(1)/(2)`

∴ equation of the normal at θ = `pi/(6), "i.e at" (1/2, 1/2)` is

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

∴ `2y - 1 = x - (1)/(2)`

∴ 4y – 2 = 2x – 1

∴ 2x – 4y + 1 = 0

Hence, equations of the tangent and normal are 4x + 2y – 3 = 0 and 2x – 4y + 1 = 0 respectively.