JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The slope of the tangent to the curve

\left(y-x^5\right)^2=x\left(1+x^2\right)^2

at the point

(1,3)

, is

View answer and explanation

Correct option: C

Option C (8) is correct because differentiating the curve equation and substituting the point (1,3) yields a slope of 8.

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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