JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The length of the normal to the curve

x = a\left( {\theta + \sin \theta } \right),y = a\left( {1 - \cos \theta } \right)at\;\theta = \frac{\pi }{2}

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Correct option: D

\sqrt {2} a

Option D (

\sqrt {2} a

) is correct because it is the calculated length of the normal to the curve at the given parameter value.

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a

y=x^3-a x

x^2 y=1-y

x y=1-y

C

y^3-3 x y+2=0

H

\mathrm{C}

x

V

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y

y-e^{x y}+x=0

y = f\left( x \right)

x

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\theta = \left( {2k + 1} \right)\pi ,\;k \in Z