Let \({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{a}}{{\rm{x}}^2} + {\rm{b}}{{\rm{y}}^2}}}{{{\rm{xy}}}}\), where a and b are constants. If \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\;}}\) at x = 1 and y = 2, then the relation between a and b is

\({\rm{a}} = \frac{{\rm{b}}}{4}\)

\({\rm{a}} = \frac{{\rm{b}}}{2}\)

Let \({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{a}}{{\r

1.	Let \({\rm{f}}\left( {{\rm{x}},{\rm{y}}} \right) = \frac{{{\rm{a}}{{\rm{x}}^2} + {\rm{b}}{{\rm{y}}^2}}}{{{\rm{xy}}}}\), where a and b are constants. If \(\frac{{\partial {\rm{f}}}}{{\partial {\rm{x}}}} = \frac{{\partial {\rm{f}}}}{{\partial {\rm{y}}}}{\rm{\;}}\) at x = 1 and y = 2, then the relation between a and b is
A.	\({\rm{a}} = \frac{{\rm{b}}}{4}\)
B.	\({\rm{a}} = \frac{{\rm{b}}}{2}\)
C.	a = 2b
D.	a = 4b
Answer» E.

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