JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Medium

1 of 1

Application +4 | -1

The normal to the curve

y\left( {x - 2} \right)\left( {x - 3} \right) = x + 6

at the point, where the curve intersects the

Y

-axis passes through the point

View answer and explanation

Correct option: B

\left( {\frac{1}{2},\frac{1}{2}} \right)

Option B (

\left( {\frac{1}{2},\frac{1}{2}} \right)

) is correct because the normal to the curve at the y-intercept passes through this point.

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