Folding an ellipse

Extend the lines and label the points as below:

Since the black line is $y = -2 - x$ and $P$ is at $x = -1$ , the coordinates for $P$ are $P(-1, -1)$ .

Let the green line have an equation of $y = mx + b$ . Then the green and black lines intersect at $Q(\frac{-b - 2}{m + 1}, \frac{b - 2m}{m + 1})$ .

$PP'$ is perpendicular to the green line and through $P(-1, -1)$ , so its equation is $y = -\frac{1}{m}(x + 1) - 1$ . It intersects the green line at $R(\frac{-1 - m - mb}{m^2 + 1}, \frac{b - m - m^2}{m^2 + 1})$ .

The coordinates for $P'$ are then $P'(P_x + 2(R_x - P_x), P_y + 2(R_y - P_y))$ or $P'(\frac{m^2 - 1 - 2m - 2mb}{m^2 + 1}, \frac{-m^2 + 1 + 2b - 2m}{m^2 + 1})$ .

The slope of $QP'$ is then $m_2 = \frac{P'_y - Q_y}{P'_x - Q_x} = \frac{-bm^2 + 2bm + b + m^3 - 3m^2 + m + 1}{-bm^2 - 2bm + b + m^3 + m^2 - 3m + 1}$ .

The ellipse with semi-major axis $5$ and semi-minor axis $3$ has an equation of $\frac{x^2}{25} + \frac{y^2}{9} = 1$ . Using implicit differentiation, the slope of the tangent line on the ellipse is $\frac{dy}{dx} = -\frac{9x}{25y}$ .

The reflected point $P'$ must be on the ellipse, so $\frac{{P'_x}^2}{25} + \frac{{P'_y}^2}{9} = 1$ or

$\frac{(\frac{m^2 - 1 - 2m - 2mb}{m^2 + 1})^2}{25} + \frac{(\frac{-m^2 + 1 + 2b - 2m}{m^2 + 1})^2}{9} = 1$

and the slope of $QP'$ must be the same as the slope of the tangent line of the ellipse at $P'$ , so $m_2 = -\frac{9P'_x}{25P'_y}$ or

$\frac{-bm^2 + 2bm + b + m^3 - 3m^2 + m + 1}{-bm^2 - 2bm + b + m^3 + m^2 - 3m + 1} = -\frac{9(\frac{m^2 - 1 - 2m - 2mb}{m^2 + 1})}{25(\frac{-m^2 + 1 + 2b - 2m}{m^2 + 1})}$

These two equations solve numerically to $(m, b) \approx (-0.479393, 0.95049)$ for $b > 0$ , which means the green line has an approximate equation of $y = -0.479393x + 0.95049$ .

The green line and the ellipse then intersect at approximately $(-3.01217, 2.3945)$ and $(4.55725, -1.23422)$ , which by the distance formula are approximately $\boxed{8.394}$ units apart.