JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The distance between the origin and the normal to the curve

y = {e^{2x}} + {x^2}\,at\,x = 0

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Correct option: C

\frac{2}{{\sqrt 5 }}

Option C (

\frac{2}{{\sqrt 5 }}

) is correct because it represents the perpendicular distance from the origin to the normal line of the given curve at x=0.

More Questions from TANGENTS & NORMALS

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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