The tangent to the curve $y-a x^2+b x$ at $(2,-8)$ is parallel to $x$-axis. Then

JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

1 of 1

Application +4 | -1

The tangent to the curve

y-a x^2+b x

(2,-8)

is parallel to

x

-axis. Then

View answer and explanation

Correct option: C

a=2, b=-8

Option C (a=2, b=-8) is correct because these values satisfy both the condition that the tangent at (2, -8) is parallel to the x-axis (dy/dx = 0) and that the point lies on the curve.

More Questions from TANGENTS & NORMALS

Practice All Chapter MCQs →

The complete set of value of '

a

' for which there exists at least one line that is tangent to the graph of the curve

y=x^3-a x

at one point and normal to the graph at another point is given by The point of intersection of the tangents drawn to the curve

x^2 y=1-y

at the points where it is met by the curve

x y=1-y

, is given by Let

C

be the curve

y^3-3 x y+2=0

. If

H

is the set of points on the curve

\mathrm{C}

where the tangent is parallel to

x

-axis and

V

is the set of points on

C

where the tangent is parallel to

y

-axis, then The curve

y-e^{x y}+x=0

has a vertical tangent at the point: If normal to the curve

y = f\left( x \right)

is parallel to

x

-axis, then The tangents to the curve

x = a\left( {\theta - \sin \theta } \right),\;y = a\left( {1 + \cos \theta } \right)

at the points

\theta = \left( {2k + 1} \right)\pi ,\;k \in Z

are parallel to: