The equation of tangent at $(-4,-4)$ on the curve $x^2=-4 y$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The equation of tangent at

(-4,-4)

on the curve

x^2=-4 y

View answer and explanation

Correct option: D

2 x-y+4=0

Option D (

2 x-y+4=0

) is the correct answer because it represents the equation of the tangent line to the curve at the given point.

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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