For a given relation \(\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} =

SOLUTION

Given equation is,

\(\sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} = P\left( {x - y} \right)\)

\(\Rightarrow \sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} - P\left( {x - y} \right) = 0\)

Consider this equation as a function of x, y

\(\Rightarrow f\left( {x,\;y} \right) = \sqrt {1 - {x^2}} + \sqrt {1 - {y^2}} - P\left( {x - y} \right)\)

For homogenous function,

\(\Rightarrow \frac{{dy}}{{dx}} = \frac{{ - \partial f/\partial x}}{{\partial f/\partial y}}\)

\(\frac{{\partial f}}{{\partial y}} = \frac{1}{{2\sqrt {1 - {y^2}} }} \cdot \left( { - 2y} \right) + P = 0\)

\(\frac{{\partial f}}{{\partial x}} = \frac{1}{{2\sqrt {1 - {x^2}} }}\left( { - x} \right) - P\) 

\(\frac{{dy}}{{dx}} = - \frac{{\left[ {\frac{{\left( { - x} \right)}}{{\sqrt {1 - {x^2}} }} - P} \right]}}{{\frac{y}{{\sqrt {1 - {y^2}} }} + P}}\)

At point (0, 0) i.e. x = 0, y = 0

\({\left. {\frac{{dy}}{{dx}}} \right|_{\left( {0,\;0} \right)}} = \frac{{ - \left[ {0 - P} \right]}}{{\left[ {0 + P} \right]}} = 1\)