Question:

Step 1: Set up the differential equation. The slope at the point $(x, y)$ is given by: $$\frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}}$$ Step 2: Integrate to find the general solution. The curve passes through $(-2, -2)$, which implies $x < -1$. For $x < -1$, let $u = -x$. Then $u > 1$. The derivative of $\sec^{-1}(u)$ with respect to $u$ is $\frac{1}{u\sqrt{u^2-1}}$. Using the chain rule, the derivative of $\sec^{-1}(-x)$ with respect to $x$ is: $$\frac{d}{dx}[\sec^{-1}(-x)] = \frac{1}{(-x)\sqrt{(-x)^2-1}} \cdot (-1) = \frac{1}{-x\sqrt{x^2-1}} \cdot (-1) = \frac{1}{x\sqrt{x^2-1}}$$ Therefore, the integral of $\frac{1}{x\sqrt{x^2-1}}$ for $x < -1$ is: $$y = \sec^{-1}(-x) + C$$ Step 3: Apply the initial condition. The curve passes through the point $(-2, -2)$. Substitute $x = -2$ and $y = -2$ into the equation: $$-2 = \sec^{-1}(-(-2)) + C$$ $$-2 = \sec^{-1}(2) + C$$ The principal value of $\sec^{-1}(2)$ is $\frac{\pi}{3}$. $$-2 = \frac{\pi}{3} + C$$ Solving for $C$: $$C = -2 - \frac{\pi}{3}$$ Thus, the equation of the curve is: $$y = \sec^{-1}(-x) - 2 - \frac{\pi}{3}$$ Step 4: Verify points on the curve. To determine other points the curve passes through, substitute the x-coordinates into the curve's equation. For $x = -\frac{2}{\sqrt{3}}$: $$-x = \frac{2}{\sqrt{3}}$$ $$y = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) - 2 - \frac{\pi}{3}$$ The principal value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is $\frac{\pi}{6}$. $$y = \frac{\pi}{6} - 2 - \frac{\pi}{3}$$ $$y = \frac{\pi - 2\pi}{6} - 2$$ $$y = -\frac{\pi}{6} - 2$$ Therefore, the curve passes through $\left(-\frac{2}{\sqrt{3}}, -\frac{\pi}{6} - 2\right)$. For $x = -\sqrt{2}$: $$-x = \sqrt{2}$$ $$y = \sec^{-1}(\sqrt{2}) - 2 - \frac{\pi}{3}$$ The principal value of $\sec^{-1}(\sqrt{2})$ is $\frac{\pi}{4}$. $$y = \frac{\pi}{4} - 2 - \frac{\pi}{3}$$ $$y = \frac{3\pi - 4\pi}{12} - 2$$ $$y = -\frac{\pi}{12} - 2$$ Therefore, the curve passes through $\left(-\sqrt{2}, -\frac{\pi}{12} - 2\right)$.