JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Application +4 | -1

The abscissae of the point on the curve

x y=(a+x)^2

, the normal at which cuts off numerically equal intercepts from the axes of coordinates, is

View answer and explanation

Correct option: A

\frac{a}{\sqrt{2}}

Option A (

\frac{a}{\sqrt{2}}

) is correct because the abscissa of the point is found by setting the negative reciprocal of the derivative equal to

\pm 1

and solving for

x

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