JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Medium

1 of 1

Application +4 | -1

(a, b)

be the point on the curve

9 y^{2}=x^{3}

where normal to the curve make equal intercepts with the axis, then the value of

(a+b)

is :

View answer and explanation

Correct option: C

\frac{20}{3}

Option C (

\frac{20}{3}

) is correct because finding the derivative, setting the normal's slope to -1, and solving the resulting equations yields the correct a and b values.

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