If the tangent at $(1,1)$ on $y^2=x(2-x)^2$ meets the curve again at $P$, then $P$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

If the tangent at

(1,1)

y^2=x(2-x)^2

meets the curve again at

P

, then

P

View answer and explanation

Correct option: C

(9 / 4,3 / 8)

Option C (

(9 / 4,3 / 8)

) is correct because the tangent line equation is found, and then solved simultaneously with the original curve to find the intersection point P.

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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