If $$\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}$$

where a > b > 0, then ${{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)$ is :

$(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}$<br><br>$$ \Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}$$<br><br>Differentiating both sides :<br><br>$$0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)$$<br><br>$$ - 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0$$<br><br>At $\left( {{\pi \over 4},{\pi \over 4}} \right)$ :<br><br>$$ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0$$<br><br>$$ \Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}$$; a, b &gt; 0