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Question

The equation of the tangent to the curve x=2cos3θ andy=3sin3θ at the pointθ=π4 is


A

2x+3y=32

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B

2x-3y=32

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C

3x+2y=32

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D

3x-2y=32

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Solution

The correct option is C

3x+2y=32


Explanation for the correct option:

Step 1: Find the first order derivative of the equation of the given curve

Given, Equation of curve x=2cos3θ and y=3sin3θ

We are given, x=2cos3θ and y=3sin3θ

dxdθ=-6cos2θsinθ and dydθ=9sin2θcosθ

dydx=dydθdxdθ=-32tanθ[sinθcosθ=tanθ]

Step 2:Find the slope of the tangent at the required point

At θ=π4,

x=2cos3π4 and y=3sin3π4

x=12 and y=322

The first order derivative at a given point gives the slope of the tangent at that point

dydxθ=π4=-32

Step 3: Find the equation of the tangent using slope-point form

Therefore, equation of tangent of the curve passing through 12,322 is

y322=-32x12y322=32x+32232x+y=322+3223x+2y2=6223x+2y=623x+2y=32

Hence, the correct answer is option (C).


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