JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The normal to the curve

5{x^5} - 10{x^3} + x + 2y + 6 = 0\,at\,P\left( {0,\; - 3} \right)

meets the curve again at the point

View answer and explanation

Correct option: B

\left( {1,\; - 1} \right),\;\left( { - 1,\; - 5} \right)

Option D (

\left( { - 1,\;5} \right),\;\left( {1,\; - 1} \right)

) is correct because by finding the equation of the normal and solving it simultaneously with the curve's equation, we find the intersection points.

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