The equation of tangent to the curve given by x = a^3 t, y = b^3 t at a point where t = ()/2 is : Option 1: x = 0 Option 2: y = 0 Option 3: y = ()/2 Option 4: x = ()/2

Answer: y = 0. Solution: At t = ()/2: x = a^3(/2) = a, y = b^3(/2) = 0. dy/dx = (-3b^2 t t)/(3a^2 t t) = -(b t)/(a t). At t = /2, slope = 0. Tangent: y = 0.

Calculus

Application of Derivatives

Medium

core

The equation of tangent to the curve given by $x = a\sin^3 t$ , $y = b\cos^3 t$ at a point where $t = \frac{\pi}{2}$ is :

$x = 0$

$y = 0$

$y = \frac{\pi}{2}$

$x = \frac{\pi}{2}$

✅ Correct Option: 2

At $t = \frac{\pi}{2}$ : $x = a\sin^3(\pi/2) = a$ , $y = b\cos^3(\pi/2) = 0$ . $\frac{dy}{dx} = \frac{-3b\cos^2 t \sin t}{3a\sin^2 t \cos t} = -\frac{b\cos t}{a\sin t}$ . At $t = \pi/2$ , slope $= 0$ . Tangent: $y = 0$ .

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The equation of tangent to the curve given by x=asin⁡3tx = a\sin^3 tx=asin3t, y=bcos⁡3ty = b\cos^3 ty=bcos3t at a point where t=π2t = \frac{\pi}{2}t=2π​ is :

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The equation of tangent to the curve given by $x = a\sin^3 t$ , $y = b\cos^3 t$ at a point where $t = \frac{\pi}{2}$ is :