If x={acos}³ θ y={asin}³ θ Then {1+≤ft(d y/d x)²}=

Correct Answer is : sec⁡θ θsecθ

Correct Answer is : sec⁡θ θsecθ

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MHTCET 22 Apr 2025 Shift 1 Paper

Question : 147 of 150

Marks: +1, -0

x=\operatorname{acos}^{3} \theta y=\operatorname{asin}^{3} \theta

\sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=

Solution:

Concept:

Parametric differentiation followed by the trigonometric identity

1 + \tan^2 \theta = \sec^2 \theta

Explanation:

Given

x = a \cos^3 \theta

and

y = a \sin^3 \theta

Differentiate

x

with respect to

\theta

\frac{dx}{d\theta} = -3a \cos^2 \theta \sin \theta

Differentiate

y

with respect to

\theta

\frac{dy}{d\theta} = 3a \sin^2 \theta \cos \theta

Now find

\frac{dy}{dx}

using the chain rule:

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} = -\tan \theta

Therefore,

\sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta

Answer:

\sec \theta

, i.e., option C.

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