Question

The equation of the tangent to the curve given by x = a sin³t, y = bcos³t at a point where t = `pi/2` is

Options

• y = 1

• y = 0

• x = 0

• x = 1

Answer 1

y = 0

Explanation:

x = a sin³t

`(dx)/(dt)` = 3a sin²t. cos t

y = b cos³t

`(dy)/(dx) = 3b cos^2t. (- sin t)`

`(dy)/(dx) = (dy/dt)/(dx/(dt)) = (-3b cos^2t. sint)/(3a sin^2t. cos t) = - b/a cot t`

Put `x = pi/2`

`m = (dy)/(dx) = 0, x = a sin^3 (pi/2) = a`

and `y = b cos^3 (pi/2)` = 0

∴ Equation of tangent  y – 0 = 0 (x – 0) ⇒ y = 0