JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Analyzing +4 | -1

The two curves

{x^3} - 3x{y^2} + 2 = 0

and

3{x^2}y - {y^3} - 2 = 0

View answer and explanation

Correct option: A

Cut at right angel

Option A (Cut at right angel) is correct because the product of the slopes of the tangents to the curves at the point of intersection is -1.

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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