JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

Point of contact of tangents to the curve

y^{2}-2 x^{3}-4 y+8=0

from the point

(1,2)

is/are :

View answer and explanation

Correct option: A

(2,2 \pm \sqrt{3})

Option A is correct because the points of contact are found by solving the tangent equation and the curve equation simultaneously.

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