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The Equation Of The Normal Line To The Curve Of X³+Y³-9xy=0 At The Point (2,4)

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Class 12

>>Maths

>>Application of Derivatives

>>Tangents and Normals

>>Find the equation of tangent and normals

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Find the equation of tangent and normals to the following curves at the indicated points on them : x

3

+y

3

−9xy=0 at (2,4)

Medium

Updated on : 2022-09-05

Solution

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x

3

+y

3

−9xy=0

Differentiating both side w.r.t. x, we get

3x

2

+3y

2

dx

dy

−9[x

dx

dy

+y.

dx

d

(x)]=0

∴3x

2

+3y

2

dx

dy

−9x

dx

dy

−9yx1=0

∴(3y

2

−9x)

dx

dy

=9y−3x

2

dx

dy

=

3y

2

−9x

9y−3x

2

∴(

dx

dy

)

at (2,4)

=

3(4)

2

−9(2)

9(4)−3(2)

2

=

48−18

36−12

=

30

24

=

5

4

=slope of the tangent at (2,4)

∴ the equation of the tangent at (2,4) is

y−4=

5

4

(x−2)

∴5y−20=4x−8

∴4x−5y+12=0

The slope of normal at (2,4)

$$=\dfrac{-1}{\left(\dfrac{dy}{dx}\right)_{at\ (2,4)}}=-\dfrac{5}

{4}$$

∴ the equation of the tangent at (2,4) is

y−4=

5

4

(x−2)

∴4y−16=−5x+10

∴5x+4y−26=0

hence, the equation of tangent and normal are

4x−5y+12=0 and 5x+4y−26=0 respectively.

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