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Question 11.3.3: Show that all points (x, y) satisfying xy = 3 lie on a level......

Show that all points (x, y) satisfying xy = 3 lie on a level curve for the function

g(x,y)={\frac{3(x y+1)^{2}}{x^{4}y^{4}-1}}
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By substituting xy = 3 in the expression for g, we find

g(x,y)=\frac{3(x y+1)^{2}}{(x y)^{4}-1}=\frac{3(3+1)^{2}}{3^{4}-1}=\frac{48}{80}=\frac{3}{5}

This shows that, for all (x, y) where xy = 3, the value of g(x, y) is a constant 3/5. Hence, any point (x, y) satisfying xy = 3 is on a level curve (at height 3/5) for g.^{5}

^{5}  In fact, {{{g(x,y)=3(c+1)^{2}/(c^{4}}-1)\ \ \text{whenever}\ x y=c\neq\pm1,}} so this equation represents a level curve for g for every value of c except c\neq\pm1.

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