Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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., $x+y=±4$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.