If $x=t^2$ and $y=2 t$, then the equation of the normal at $t=1$ is

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

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Application +4 | -1

x=t^2

and

y=2 t

, then the equation of the normal at

t=1

View answer and explanation

Correct option: A

x+y-3=0

Option A (

x+y-3=0

) is correct because by finding the derivative and the slope of the normal, we can determine the equation of the normal at the given point.

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