Question

Find the tangent and normal to the following curves at the given points on the curve

x = cos t, y = 2 sin²t at t = `pi/2`

Answer 1

x = cos t, y = 2 sin²t at t = `pi/2`

At t = `pi/3`, x= cos  `pi/3 = 1/2`

At t = `pi/3`, y = `2sin^2  pi/3 = 2(3/4) = 3/2`

Point is `(1/2, 3/2)`

Now x = cos t y = 2 sin²t

Differentiating w.r.t. ‘t’,

`("d"x)/("d"y) = - sin "t"`

`("d"y)/"dt"` = 4 sin t cos t

Slope of the tangent

m = `("d"y)/("d"x)`

= `(("d"y)/("dt"))/(("d"x)/("dt"))`

= `(4 sin "t" cos "t")/(- sin "t")`

= – 4 cos t

`(("d"y)/("d"x))_(("t" = pi/3)) = - 4 cos  pi/3 = - 2`

Slope of the Normal `- 1/"m" = 1/2`

Equation of tangent is

y – y₁ = m(x – x₁)

⇒ `y - 3/2 = - 2(x - 1/2)`

⇒ 2y – 3 = – 4x + 2

⇒ 4x + 2y – 5 = 0

Equation of Normal is

`y - y_1 = - 1/"m"(x - x_1)`

⇒ `y - 3/2 = 1/2(x - 1/2)`

⇒ 2(2y – 3) = 2x – 1

⇒ 4y – 6 = 2x – 1

⇒ 2x – 4y + 5 = 0