Call for coordinate?

Let us take the point P on the curve $y = f(x)$ as $P(x_0,f(x_0))$ with its normal line expressible as:

$y - f(x_0) = (\frac{-1}{f'(x_0)})*(x - x_0) \Rightarrow y = \frac{-1}{f'(x_0)}x + [f(x_0) + \frac{x_0}{f'(x_0)}]$ (i).

and the perpendicular distance line from the origin to (i) be:

$y = f'(x_0)x$ (ii).

Equating (i) with (ii) gives the intersection point Q:

$Q(x,y) = (\frac{x_0 + f(x_0)f'(x_0)}{f'(x_0)^2 + 1}, \frac{x_0f'(x_0) + f(x_0)f'(x_0)^2}{f'(x_0)^2 + 1})$ (iii).

Now, the distance from Q to the origin equals the distance of P to the x-axis, or $|f(x_0)|$ . This can be expressed via the square of the 2-D distance formula:

$f(x_0)^2 = (\frac{x_0 + f(x_0)f'(x_0)}{f'(x_0)^2 + 1})^2 + (\frac{x_0f'(x_0) + f(x_0)f'(x_0)^2}{f'(x_0)^2 + 1})^2$ . (iv)

which after expanding and simplifying both sides of (iv) will give:

$f(x_0)^2[f'(x_0)^2 + 1] = x_0^2 + 2x_0f(x_0)f'(x_0) + f(x_0)^2f'(x_0)^2 \Rightarrow f(x_0)^2 = x_0^2 + 2x_0f(x_0)f'(x_0)$ (v).

or $\frac{y^2 - x^2}{2xy} = \frac{y}{2x} - \frac{x}{2y} = \frac{dy}{dx}$ . Now make the substitution $v = \frac{y}{x}, \frac{dy}{dx} = v + x\frac{dv}{dx}$ to obtain:

$v + x\frac{dv}{dx} = \frac{v}{2} - \frac{1}{2v} \Rightarrow x\frac{dv}{dx} = \frac{v^2 -1}{2v} - \frac{2v^2}{2v} \Rightarrow x\frac{dv}{dx} = -\frac{v^2 + 1}{2v}$ .

We can now obtain the separable first-order ODE $-\frac{dx}{x} = \frac{2v dv}{v^2 + 1}$ , which integrates into $-ln(x) + C = ln(v^2 + 1) \Rightarrow -ln(x) + C = ln(\frac{y^2}{x^2} + 1)$ . Plugging in the boundary condition $y = f(1) = 1$ yields $C = ln(2)$ , or:

$ln(\frac{2}{x}) = ln(\frac{y^2}{x^2} + 1) \Rightarrow \frac{2}{x} = \frac{y^2 + x^2}{x^2} \Rightarrow 0 = x^2 + y^2 - 2x$ (vi)

with $a = b = 1, c = f = h = 0, 2g = -2$ . Hence the required coefficient sum is $a + b + 2g = 1 + 1 - 2 = \boxed{0}$ .