JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Hard

1 of 1

Application +4 | -1

The length of normal to the curve

x=a(\theta+\sin \theta)

y=a(1-\cos \theta)

at the point

\theta=\pi / 2

View answer and explanation

Correct option: C

\sqrt{2} a

Option C (

\sqrt{2} a

) is the correct answer because it is the calculated length of the normal to the curve at the specified point.

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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

C

y

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

y = f\left( x \right)

x

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

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