Hyperbolas and Triangle!

Since $H_1$ is $y = \frac{4}{x}$ , the slope at any point $x$ is given by $m_1 = -\frac{4}{x^2}$ . Similarly, since $H_2$ is $y = \frac{1}{x}$ , the slope at any point $x$ is given by $m_2 = -\frac{1}{x^2}$ .

Let $P$ on $H_1$ be $(p, \frac{4}{p})$ . Then its slope is $m_P = -\frac{4}{p^2}$ , and its tangent line equation is $(y - \frac{4}{p}) = -\frac{4}{p^2}(x - p)$ or $y = -\frac{4}{p^2}x + \frac{8}{p}$ .

This tangent line intersects $H_2$ , or $y = \frac{1}{x}$ , when $y = \frac{1}{x} = -\frac{4}{p^2}x + \frac{8}{p}$ , which solves to $x = \frac{p}{2}(2 \pm \sqrt{3})$ , and then $y = \frac{2}{p}(2 \mp \sqrt{3})$ . So $A$ is $(\frac{p}{2}(2 + \sqrt{3}), \frac{2}{p}(2 - \sqrt{3}))$ and $B$ is $(\frac{p}{2}(2 - \sqrt{3}), \frac{2}{p}(2 + \sqrt{3}))$ .

The slope at $A$ on $H_2$ is $m_A = -\frac{1}{(\frac{p}{2}(2 + \sqrt{3}))^2}$ , and its tangent line calculates to $y = -\frac{4}{p^2}(7 - 4\sqrt{3})x + \frac{4}{p}(2 - \sqrt{3})$ . The slope at $B$ on $H_2$ is $m_B = -\frac{1}{(\frac{p}{2}(2 - \sqrt{3}))^2}$ , and its tangent line calculates to $y = -\frac{4}{p^2}(7 + 4\sqrt{3})x + \frac{4}{p}(2 + \sqrt{3})$ . Then $y = -\frac{4}{p^2}(7 - 4\sqrt{3})x + \frac{4}{p}(2 - \sqrt{3}) = -\frac{4}{p^2}(7 + 4\sqrt{3})x + \frac{4}{p}(2 + \sqrt{3})$ , which solves to $x = \frac{p}{4}$ , and then $y = \frac{1}{p}$ . So $Q$ is $(\frac{p}{4}, \frac{1}{p})$ .

This means vector $QA = (\frac{p}{4}(3 + 2\sqrt{3}), \frac{1}{p}(3 - 2\sqrt{3}))$ and vector $QB = (\frac{p}{4}(3 - 2\sqrt{3}), \frac{1}{p}(3 + 2\sqrt{3}))$ , and the area of $\triangle QAB = \frac{1}{2} \begin{vmatrix}\frac{p}{4}(3 + 2\sqrt{3}) & \frac{1}{p}(3 - 2\sqrt{3}) \\ \frac{p}{4}(3 - 2\sqrt{3}) & \frac{1}{p}(3 + 2\sqrt{3}) \end{vmatrix}$ $=$ $3\sqrt{3} \approx \boxed{5.196152423}$ .

Since the two curves are hyperbolas, the area of any such triangle will be the same. So just pick a point on the y=4/x graph and calculate the slope of that point. For simplicity, I picked the point (2,2) with slope of -1. (f'(x)=-4x^-2 , then plug in 2 for x). If you extend the tangent line to the y axis it's easy to see that the equation of that tangent line is y=-x+4. Set that equal to the other hyperbola to find points A and B: (1/x)=-x+4 The solutions are x= 2+sqrt(3) and x=2-sqrt(3) Since it's a hyperbola, the y coordinates will be the conjugate of those numbers. Now all is left is to find the derivatives at those points, find the equations for those lines, and set them equal to each other. Providing you can navigate basic calculus and radicals, the coordinates of Q should be roughly x=.4948 y=.5724. I had rounding errors but it's close enough. There's a fancy calculator online where it does the distance formula and the area formula for SSS triangles. AREA= 5.1 plus or minus .1