JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

At what point on the curve

{x^3} - 8{a^2}y = 0

, the slope of the normal is

- \frac{2}{3}

View answer and explanation

Correct option: C

\left( {2a,\;a} \right)

Option D (

\frac{x}{a} + \frac{y}{b} = 1

) is correct because it represents the equation of the tangent to the curve at the point where it crosses the y-axis.

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