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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`
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Solution
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
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