JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

If the tangent to the curve

2 y^3=a x^2+x^3

at the point

(a, a)

cuts off intercepts

\alpha

and

\beta

on the coordinate axes such that

\alpha^2+\beta^2=61

, then

a=

View answer and explanation

Correct option: A

\pm 30

Option A (

\pm 30

) is correct because substituting the given conditions into the equation and its derivative allows us to solve for a.

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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