If $x\sqrt {1-y^2} + y\sqrt {1-x^2}=a$, show that

Question

If $x\sqrt {1-y^2} + y\sqrt {1-x^2}=a$, show that $\dfrac {d^2y}{dx^2} =-\dfrac {a}{(1-x^2)^{\frac {3}{2}}}$

I thought of using implicit differentiation, but it needs to be used twice which is cumbersome and complex. Isn't there any idea?

Answer 1

HINT:

WLOG $x=\sin A,y=\sin B$ where $\dfrac\pi2\le A,B\le\dfrac\pi2$

$$\implies\sin(A+B)=a$$

$$ \arcsin a=\arcsin x+\arcsin y$$

Answer 2

solving your equation for $y$ we get $$y=a\sqrt{1-x^2)}-\sqrt{x^2-a^2x^2}$$ or $$y=\sqrt{x^2-a^2x^2}+a\sqrt{1-x^2}$$ then you can differentiate this with respect to $x$