JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Hard

1 of 1

Application +4 | -1

The sum of intercepts on the coordinate axes made by the tangent to the curve

\sqrt{x}+\sqrt{y}=\sqrt{a}

View answer and explanation

Correct option: A

a

Option A (

a

) is correct because the sum of the intercepts of the tangent to the curve is equal to

a

, as derived from the given equation.

More Questions from TANGENTS & NORMALS

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