Find the angle between $x^2+y^2=8$ and $xy=4$ at the intersection points. I have thought in find the angle between the tangent lines of each function at the intersection point, but i don´t know how to do it


Let the intersection point be $(a,b)$, so $a^2+b^2=8$ and $ab=4≠0$ $$\implies \frac{a^2+b^2}{ab}=\frac{4}{2}\implies (a-b)^2=0\implies a=b≠0$$

So,$a=±2$ as $ab=4$.

$$x^2+y^2=4\implies 2x+2y\frac{dy}{dx}=0\implies (\frac{dy}{dx})_{x=y}=-1$$

$$xy=4\implies x\frac{dy}{dx}+y=0\implies (\frac{dy}{dx})_{x=y}=-1 $$

So, at $(a,a)$ the gradients of the two given curves are same.

We know if the angle between the two curves is $\theta$, then $\tan\theta=\frac{g_1-g_2}{1+g_1g_2}$ where $g_i$ s are the gradient of the intersecting curves.

So , the angle between the given two curves at $(a,a)$, i.e., at $(2,2)$ and $(-2,-2)$, is $$\tan^{-1}\left(\frac{-1-(-1)}{1+(-1)(-1)}\right)=\tan^{-1}(0)=0 \text{ or }\pi$$

Alternatively, we are given $x^2+y^2=8$ and $xy=4$. From the latter, $x=\frac{4}{y}$

$$\left(\frac{4}{y}\right)^2+y^2=8\implies y^4-8y^2+16=0\implies (y^2-4)^2=0\implies y=±2$$

$$\tag 1 y=2\implies x=2$$ and $$\tag 2 y=-2\implies x=-2$$

So, $x=y$ at the points of intersection.

Observe that the equation of the common tangents are $x+y=a+a=2a$ i.e., $x+y=±4$.


Suppose that near the intersection point the first curve is parametrized by $y_1(x)$ and the second by $y_2(x)$. Plugging in $y = y_1(x)$ and differentiating w.r.t. $x$ we get $$2x + 2y_1(x)y_1'(x) = 0$$ $$y_1'(x) = - x / y_1(x).$$ Similarly for the second curve we get $$y_2'(x) = -y_2(x) / x.$$

Now, to find intersection points we subtract from the first curve equation twice the second one to find $$x^2 + y^2 - 2xy = (x - y)^2 = 0.$$ Therefore $y = x$ and from the second equation $x^2 = 4$, so that the intersection points are $(2, 2)$ and $(-2, -2)$. From the above discussion we have for the slope of $y_1$ is $y_1'(2) = -2 / 2 = -1$ and similarly $y_2'(2) = -1$. Since they are equal, the curves are tangent and the sought for angle is zero.


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