JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

Let slope of the tangent line to a curve at any point

P\,(x,\,y)

be given by

\frac{{x{y^2} + y}}{x}

. If the curve intersects the line

x + 2y = 4

x = - 2

, then the value of

y

, for which the point

(3,\,y)

lies on the curve, is

View answer and explanation

Correct option: C

- \frac{{18}}{{19}}

Option C (

- \frac{{18}}{{19}}

) is correct because it is the value of y that satisfies the differential equation and the given condition.

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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