JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Analyzing +4 | -1

The point of intersection of the tangents drawn to the curve

x^2 y=1-y

at the points where it is met by the curve

x y=1-y

, is given by

View answer and explanation

Correct option: C

(0,1)

Option C ((0,1)) is correct because it is the point of intersection of the tangents to the curve at the points where it intersects another curve.

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y=x^3-a x

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y^3-3 x y+2=0

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\mathrm{C}

x

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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

{x^2} = 4\left( {y + 1} \right)

{x^2} = - 4\left( {y + 1} \right)