JEE MATHEMATICS MCQ: TANGENTS & NORMALS

1 Easy

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Application +4 | -1

The equation of the normal to the curve

y=e^{-2|x|}

at the point where the curve cuts the line

x=-\frac{1}{2}

, is

View answer and explanation

Correct option: A

2 e(e x+2 y)=4-e^2

Option A (

2 e(e x+2 y)=4-e^2

) is correct because it represents the equation of the normal to the curve at the given point, derived using the derivative and the negative reciprocal of the slope.

More Questions from TANGENTS & NORMALS

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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: