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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.