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The equation of normal at any point $θ$ to the curve $x = acos θ + a θsin θ, y = asin θ - a θcos θ$ is always at a distance of

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2a units from origin

a units from origin

\frac{1}{2} a

units from origin

5a units from origin

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detailed solution

Correct option is B

$\begin{array}{l} x = acos θ + aθsin θ, y = asin θ - aθcos θ \\ \frac{dy}{dx} = \frac{aθsin θ}{aθcos θ} = \frac{\sin θ}{\cos θ} \end{array}$

Slope of normal $= \frac{- \cos θ}{\sin θ}$

Equation of normal at $θ$

$\begin{array}{l} y - [asin θ - aθcos θ] = \frac{- \cos θ}{\sin θ} [x - (acos θ + aθsin θ)] \\ ysin θ - {asin}^{2} θ + aθsin θcos θ = - xcos θ + {acos}^{2} θ + aθsin θcos θ \\ xcos θ + ysin θ - a = 0 \end{array}$

The distance from origin to the normal

$= \frac{| - a |}{\sqrt{\cos^{2} θ + \sin^{2} θ}} = a$

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The equation of normal at any point θ to the curve x=acos⁡θ+a θsin⁡θ, y=asin⁡θ−a θcos⁡θ is always at a distance of

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The equation of normal at any point $θ$ to the curve $x = acos θ + a θsin θ, y = asin θ - a θcos θ$ is always at a distance of