Let $y = f(x)$ be a curve $C_1$ passing through $(2,2)$ and $\left(8, \frac{1}{2}\right)$ and satisfying a differential equation $y\left(\frac{d^2y}{dx^2}\right) = 2\left(\frac{dy}{dx}\right)^2$. Curve $C_2$ is the director circle of the circle $x^2 + y^2 = 2$. If the shortest distance between the curves $C_1$ and $C_2$ is $\left(\sqrt{p-q}\right)$ where $p,q \in \mathbb{N}$, then find the value of $\left(p^2 - q\right)$.
Step-by-Step Solution
Key Concept: Solve the differential equation using substitution $p = \frac{dy}{dx}$ to get a hyperbola, then find its minimum distance to the director circle $x^2 + y^2 = 4$.
To solve the differential equation $y\frac{d^2y}{dx^2} = 2\left(\frac{dy}{dx}\right)^2$, let $p = \frac{dy}{dx}$, giving $yp\frac{dp}{dy} = 2p^2$. For $p \neq 0$, this simplifies to $y\frac{dp}{dy} = 2p$, yielding $\frac{dp}{p} = \frac{2dy}{y}$. Integrating: $\ln p = 2\ln y + c_1$, so $p = Ay^2$ where $A$ is constant. Thus $\frac{dy}{dx} = Ay^2$, integrating gives $-\frac{1}{y} = Ax + B$. Using boundary conditions $(2,2)$ and $(8,\frac{1}{2})$: $-\frac{1}{2} = 2A + B$ and $-2 = 8A + B$. Solving yields $A = -\frac{1}{4}$, $B = 0$, so $y = \frac{4}{4-x}$, which is hyperbola $C_1$. The director circle of $x^2 + y^2 = 2$ is $C_2: x^2 + y^2 = 4$ (circle with radius $2$). The shortest distance between $C_1$ and $C_2$ is found by minimizing distance from origin to $C_1$. The minimum distance from origin to hyperbola $y = \frac{4}{4-x}$ is $\sqrt{2}$, so shortest distance to circle is $2 - \sqrt{2} = \sqrt{6-4} = \sqrt{p-q}$ with $p=6, q=4$.
Correct Answer: 62