Did you say scale it?

I proceeded in three steps. Having limited time, I "cheated" and left the numerical work in Step 1 to the computer.

First find a line $x=a$ that cuts the area of the unit circle in a ratio of 2:1. Solving $\int_{0}^{a} \sqrt{1-x^2}dx=\frac{1}{2}(a\sqrt{1-a^2}+\arcsin a)=\frac{\pi}{12}$ numerically gives $a\approx 0.26493208$ .

Second, find a line $y=\frac{6}{7}x+b$ that cuts the area of the unit circle in a ratio of 2:1. We want the line's distance from the origin to be $a$ , so $b=a\sqrt{1+(\frac{6}{7})^2}$ .

Finally, we are scaling our line by a factor of 12 horizontally and by a factor of 7 vertically, $x'=12x$ and $y'=7y$ , to take the form $y'=\frac{x'}{2}+7b$ with $y_0=7b=\sqrt{85}a \approx 2.4425531$ .

The required answer is $\boxed{244255}$

We start off by scaling the ellipse into the unit circle, by defining the image of $(x, y)$ to be $(x', y')$ with $x' = \dfrac{x}{a}$ ,and $y' = \dfrac{y}{b}$ .

The image of the line becomes: $b y' = \frac{1}{2} a x' + y_0$

Now this line is to be distant from the origin a distance $D$ , such that the area of the circle above $y = D$ is $\frac{1}{3}$ of the unit circle area.

Let $\theta = 2 \cos^{-1} (D)$ , then the upper area is $= \frac{1}{2} ( \theta - \sin(\theta) )$ .

Hence , $\frac{1}{3} \pi = \frac{1}{2} (\theta - \sin(\theta) ) = \frac{1}{2} ( 2 \cos^{-1}(D) - \sin(2 \cos^{-1}(D) ) )$

We can find $D$ numerically, by a suitable numerical method, such as Newton's method. To use this method, we

define the function $f(x) = \frac{1}{2} ( 2 \cos^{-1}(x) - \sin(2 \cos^{-1}(x) ) ) - \frac{1}{3} \pi$ , and we want the root of this function.

Newton's iteration, is given by $x_{n+1} = x_{n} - \dfrac{f(x_n)}{f'(x_n)}$ , where the derivative of our function is $f'(x) = \frac{1}{2} ( \dfrac{-2}{\sqrt{1 - x^2}} - \cos(2 \cos^{-1}(x) ) \cdot \dfrac{(-2)}{\sqrt{1 - x^2}} )$

This can be simplified and re-arranged into $f'(x) = - 2 \sqrt{1 - x^2}$

Putting it all together and using any programming environment or a standard spreadsheet like Microsoft Excel, we can implement the Newton Iteration. I have taken the initial guess $x_0 = 0.5$ . You'll note that the iterations converge very quickly to our sought solution and we have the desired solution $D = 0.264932085$

Now we are ready to continue the solution. Recall that our line is given by $b y' = \frac{1}{2} a x' + y_0$

Its distance from the origin is given by: $D = \dfrac{y_0}{ \sqrt{ b^2 + \frac{a^2} {4} }}$

Thus $y_0 = D \sqrt{b^2 + \frac{a^2} { 4}} = 0.264932085 \sqrt{49 + \frac{144}{4}} = 2.44255313$

This makes the answer $\lfloor 10^5 y_0 \rfloor = \boxed{244255}$