JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Hard

1 of 1

Analyzing +4 | -1

The points of contact of the tangents drawn from the origin to the curve

y = \sin x

lie on the curve

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because it represents the locus of the points of contact of tangents drawn from the origin to the curve

y = \sin x

More Questions from TANGENTS & NORMALS

Practice All Chapter MCQs →

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