Find the radius of the curvature at the origin on the curve x 3 + y 3 = 3axy
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Answer:
Given, x
2
+y
3
=3axy...(1)
Differentiating equation,
⇒3x
2
+3y
2
dx
dy
=3a(x
dx
dy
+y)
⇒3x
2
+3y
2
dx
dy
=3ax
dx
dy
+3ay
⇒(3y
2
−3ax)
dx
dy
=3ay−3x
2
⇒3(y
2
−ax)
dx
dy
=2(ay−x
2
)
⇒
dx
dy
=
y
2
−ax
ay−x
2
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