JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

The point

P

of the curve

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

such that the tangent at P is perpendicular to the line

4x - 3y + 2 = 0

is given by

View answer and explanation

Correct option: D

\left( {1/8,\; - 1/16} \right)

Option D ((1/8, -1/16)) is correct because the tangent at this point is perpendicular to the given line.

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