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The maximum distance from origin of a point on the curve x = asint-bsin(atb), y = acost-bcos(atb), both a, b > 0 is ______. -

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Question

The maximum distance from origin of a point on the curve x = `a sin t - b sin((at)/b)`, y = `a cos t - b cos((at)/b)`, both a, b > 0 is ______.

Options

  • a – b

  • a + b

  • `sqrt(a^2 + b^2)`

  • `sqrt(a^2 - b^2)`

MCQ
Fill in the Blanks

Solution

The maximum distance from origin of a point on the curve x = `a sin t - b sin((at)/b)`, y = `a cos t - b cos((at)/b)`, both a, b > 0 is a + b.

Explanation:

We know that distance of origin from

(x, y) = `sqrt(x^2 + y^2)`

= `sqrt(a^2 + b^2 - 2ab cos(t - (at)/b)`;

`≤ sqrt(a^2 + b^2 + 2ab)[{cos(t - (at)/b)}_min = -1]`

= a + b

∴ Maximum distance from origin = a + b

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