# If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, prove that $\frac{dy}{dx}=\sqrt{\frac{1-y^2}{1-x^2}}$

My Attempt $$\frac{-2x}{2\sqrt{1-x^2}}-\frac{2y}{2\sqrt{1-y^2}}.\frac{dy}{dx}=a-a\frac{dy}{dx}\\ \implies \frac{dy}{dx}\bigg[a-\frac{y}{\sqrt{1-y^2}}\bigg]=a+\frac{x}{\sqrt{1-x^2}}\\ \frac{dy}{dx}=\frac{a\sqrt{1-x^2}+x}{\sqrt{1-x^2}}.\frac{\sqrt{1-y^2}}{a\sqrt{1-y^2}+x}=\sqrt{\frac{1-y^2}{1-x^2}}.\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}$$

How do I poceed further and find the derivative ?

• you can eliminate $a$ with your first equation! – Dr. Sonnhard Graubner Apr 5 '18 at 10:26

After my comment you will get $$a\sqrt{1-x^2}+x=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and $$a\sqrt{1-y^2}-y=\frac{1-xy+\sqrt{1-x^2}\sqrt{1-y^2}}{x-y}$$ and you will get the desired result!

• How did you find this! I liked both your and JJ's answer, but how you found this? – King Tut Apr 5 '18 at 10:45
• Hm, intuition and the result containes no $a$, these are the reasons – Dr. Sonnhard Graubner Apr 5 '18 at 10:51

Both $x,y\in[-1,1]$. So, $x=\sin\theta$ and $y=\sin\phi$ for some $\displaystyle \theta,\phi\in\left[\frac{-\pi}{2},\frac{\pi}{2}\right]$. Note that $\cos\theta, \cos\phi\ge0$.

We have $$\cos\theta+\cos\phi=a(\sin\theta-\sin\phi)$$

So, $$2\cos\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2}=2a\cos\frac{\theta+\phi}{2}\sin\frac{\theta-\phi}{2}$$

$\displaystyle \tan\frac{\theta-\phi}{2}=\frac{1}{a}$ is a constant. So, $\displaystyle \frac{d\phi}{d\theta}=1$.

$$\frac{dy}{dx}=\frac{\cos\phi}{\cos\theta}\frac{d\phi}{d\theta}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$$

• you beat me to it ! +1 – David Quinn Apr 5 '18 at 10:49
• Good answer but how can we say $y\in [0,1]$? $y\in [-1,0]$ according to plot for positive $a$ – King Tut Apr 5 '18 at 10:56
• @$y\in[-1,1]$. But $\cos\phi=\sqrt{1-y^2}\in[0,1]$. This is true as $\phi \in[\frac{-\pi}{2},\frac{\pi}{2}]$. – CY Aries Apr 5 '18 at 11:25
• I have a typo here. Will edit it – CY Aries Apr 5 '18 at 11:32
• Please, take a look to my solution, in the spirit of yours. – Jean Marie Nov 14 '19 at 8:00

HINT : There is no parameter $a$ in the formula to be proved. So, first transform the initial equation into an equation where $a$ will be immediately eliminated by differentiation: $$\frac{\sqrt{1-x^2}+\sqrt{1-y^2}}{x-y}=a$$ Differentiate and simplify.

• @KingTut just do the differentiation from the form of the equation stated by@JJacquelin u'll get the final answer. So my original post is certainly not incorrect. – ss1729 Apr 5 '18 at 10:39
• @ss ok let me check the error. thanks! – King Tut Apr 5 '18 at 10:40

Move $a(x-y)$ to the LHS to make $F(x,y)=0$. Then: $$\frac{dy}{dx}=-\frac{F'_x}{F'_y}=-\frac{-\frac{x}{\sqrt{1-x^2}}-a}{-\frac{y}{\sqrt{1-y^2}}+a}=\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}\cdot \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}},$$ because: \begin{align}&\frac{a\sqrt{1-x^2}+x}{a\sqrt{1-y^2}-y}= \\ &1 \iff a(\sqrt{1-x^2}-\sqrt{1-y^2})=-(x+y) \iff a(y^2-x^2)= \\ &-(x+y)(\sqrt{1-x^2}+\sqrt{1-y^2}) \iff a(x-y)=\\ &\sqrt{1-x^2}+\sqrt{1-y^2}.\end{align}

Let us remark that the looked for differential equation can be written under the form

$$\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\tag{1}$$

involving solutions of the form :

$$-\arccos(x)=-\arccos(y)+a\tag{2}$$

otherwise said with cartesian equation :

$$y=\cos(\arccos(x)-a)\tag{3}$$

where $$a$$ is any real.

Setting $$\alpha=\arccos(x)$$ in (3), we get the equivalent parametric equations :

$$\begin{cases}x&=&\cos(\alpha)\\y&=&\cos(\alpha-a)\end{cases}\tag{4}$$

in which we recognize that we are working with elliptical arcs, as shown on the graphical representation of the family of curves $$C_a$$ displayed below (we have to pay attention to the domains of variables $$x$$ and $$y$$ : in general we will not have the whole ellipses as solutions but arcs of them).

From (4), it is easy to establish a connection with relationship:

$$\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$$

(in the spirit of the solution given by @CY Aries).

Fig. 1:Curves $$C_a$$ for $$a=-\pi$$ to $$a=\pi$$ with step $$\pi/8$$ (progressively changing from blue to red). The curves are elliptical arcs with two degenerate cases (straight lines).