JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of normal at any point

(x,y),x > 0,y > 0

on the curve

y = y(x)

is given by

\frac{{{x^2}}}{{xy - {x^2}{y^2} - 1}}.

If the curve passes through the point (1, 1) then

e.y(e)

is equal to

View answer and explanation

Correct option: D

\frac{{1 + \tan (1)}}{{1 - \tan (1)}}

Option D (

\frac{{1 + \tan (1)}}{{1 - \tan (1)}}

) is correct because by solving the differential equation and applying the initial condition, we can find the value of

e.y(e)

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