JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Application +4 | -1

For which of the following curves, the line

x + \sqrt 3 y = 2\sqrt 3

is the tangent at the point

\left( {\frac{{3\sqrt 3 }}{2},\,\,\,\,\frac{1}{2}} \right)

View answer and explanation

Correct option: D

{x^2} + 9{y^2} = 9

Option D (

{x^2} + 9{y^2} = 9

) is the correct curve because the given line is tangent to it 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