X And Y Flying About

Expanding the given equation with a common denominator, we have $2\cdot \frac{x^2 +y^2}{x^2-y^2} = 1,$ so $2 + \frac{1}{2} = \frac{5}{2} =\frac{x^2-y^2}{x^2+y^2} + \frac{x^2+y^2}{x^2-y^2} = 2\cdot \frac{x^4+y^4}{x^4-y^4}.$ Thus, the answer is $\frac{5/2}{2}+\frac{2}{5/2} = \frac{41}{20},$ so $41+20=61.$

I really appreciate Sir Eli Ross's solution becuase it very perfect.

My solution is here-

We can rewrite the $2$ nd equation as

$\huge\ \frac { { (x + y) }^{ 2 } + { (x - y) }^{ 2 } }{ { x }^{ 2 } - { y }^{ 2 } } = 1$

i.e., $x = y\sqrt { 3 }$ .

Putting this in the equation whose value is to be found, we get

$\huge\ \frac { 9{ y }^{ 4 } + { y }^{ 4 } }{ 9{ y }^{ 4 } - { y }^{ 4 } } + \frac { 9{ y }^{ 4 } - { y }^{ 4 } }{ 9{ y }^{ 4 } + { y }^{ 4 } } = \frac { 10{ y }^{ 4 } }{ 8{ y }^{ 4 } } + \frac { 8{ y }^{ 4 } }{ 10{ y }^{ 4 } } = \frac { 5 }{ 4 } + \frac { 4 }{ 5 } = \frac { 41 }{ 20 } = \frac {a}{b}$ .

Thus $a + b = \boxed{61}$ .