Question

The equation of the tangent to the curve x = 2 cos³ θ and y = 3 sin³ θ at the point θ = `pi/4` is ______.

विकल्प

• `2x + 3y = 3sqrt(2)`

• `2x - 3y = 3sqrt(2)`

• `3x + 2y = 3sqrt(2)`

• `3x - 2y = 3sqrt(2)`

Answer 1

The equation of the tangent to the curve x = 2 cos³ θ and y = 3 sin³ θ at the point θ = `pi/4` is `3x + 2y = 3sqrt(2)`.

Explanation:

At `theta = pi/4`,

x = `2 cos^3  pi/4 = 1/sqrt(2)` and y = `3sin^3  pi/4 = 3/(2sqrt(2))`

x = `2cos^3theta` and y = `3sin^3theta`

∴ `("d"x)/("d"theta) = -6 cos^2theta sin theta` and `("d"y)/("d"theta) = 9 sin^2theta cos theta`

∴ `("d"y)/("d"x) = (("d"y)/("d"theta))/(("d"x)/("d"theta)) = - 3/2 tan theta`

∴ `(("d"y)/("d"x))_((theta = pi/4)) = - 3/2`

∴ equation of the tangent at `(1/sqrt(2), 3/(2sqrt(2)))` is `y - 3/(2sqrt(2)) = - 3/2(x - 1/sqrt(2))`

⇒ `3x + 2y = 3sqrt(2)`