Prove derivative of an implicit function

Question

The question: $\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$ prove that $\dfrac{dy}{dx} = \sqrt{\dfrac{1+y^2}{1+x^2}}$

I tried doing it the normal way and got $\dfrac{dy}{dx} = \dfrac{(a\sqrt{1+x^2}-x)(\sqrt{1+y^2})}{(a\sqrt{1+y^2} + y)(\sqrt{1+x^2})}$

I also tried doing a trigonometry sub to get a relation of $a$ into a independent constant
$x=\tan(\alpha)$
$y=\tan(\beta)$

And got $$a = \dfrac{\cos(\alpha) + \cos(\beta)}{\sin(\alpha - \beta)}$$ I don't know if this one can be further manipulated.

Is there something I missed? or is there another way of solving it?

Answer 1

Need to solve it by using trig substituion and complex number $$\sqrt{1+x^2} + \sqrt{1+y^2} = a(x-y)$$ can be written as
$$\sqrt{1-(ix)^2} + \sqrt{1-(iy)^2} = -ai(ix-iy)$$ here the function is defined as $0 \leq (ix)^2 \leq 1$ and so for $(iy)^2$ $$ix = \sin(\alpha) \tag{i}$$ $$iy = \sin(\beta)\tag{ii}$$ So the equation will be now $$\cos(\alpha) + \cos(\beta) = -ai(\sin(\alpha) -\sin(\beta))$$ using $\cos(C) + \cos(D)$ and $\sin(C) - \sin(D)$ \begin{align*} \cos(\dfrac{\alpha - \beta}{2})&= -ai(\sin(\dfrac{\alpha-\beta}{2}))\\ \dfrac{\alpha - \beta}{2} &= \text{arccot}(-ai)\\ \alpha - \beta &= 2\text{arccot}(-ai) \end{align*} from (i) & (ii) $$\arcsin(ix) - \arcsin(iy) = 2\text{arccot}(-ai)$$ differentiating both sides \begin{align*}\dfrac{i}{\sqrt{1+x^2}} - \dfrac{i\dfrac{dy}{dx}}{\sqrt{1+y^2}} &= 0\\ \dfrac{dy}{dx} &= \sqrt{\dfrac{1+y^2}{1+x^2}} \end{align*}

Answer 2

Note that $$a=\frac{\sqrt{1+x^{2}}+\sqrt{1+y^{2}}}{x-y}.$$

so that $$ \begin{align*} \frac{a\sqrt{1+x^{2}}-x}{a\sqrt{1+y^2}+y} &=\frac{\frac{(1+x^{2})+\sqrt{1+x^{2}}\sqrt{1+y^{2}}-x^2+xy}{x-y}}{\frac{(1+y^{2})+\sqrt{1+x^{2}}\sqrt{1+y^2}+xy-y^{2}}{x-y}}\\ &=\frac{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}{1+\sqrt{1+x^{2}}\sqrt{1+y^{2}}+xy}\\ &=1. \end{align*}$$ Therefore from your own calculations $$\frac{dy}{dx}=\frac{(a\sqrt{1+x^{2}}-x)(\sqrt{1+y^{2}})}{(a\sqrt{1+y^{2}}+y)(\sqrt{1+x^{2}})}=\frac{\sqrt{1+y^{2}}}{\sqrt{1+x^{2}}}.$$