Eccentricity Relation for Intersection of Parabola and Ellipse Latus Recta

To find the value of $e^2 + 2\sqrt{2}$, we analyze the equations and properties of the given parabola and ellipse.

1. Latus Rectum of the Parabola $P$: The equation of the parabola is given by: $P : y^2 = 4kx$ The focus of this parabola is at $(k, 0)$, and its latus rectum is the vertical line segment passing through the focus with endpoints at: $(k, 2k) \quad \text{and} \quad (k, -2k)$

2. Latus Rectum of the Ellipse $E$: The equation of the ellipse is given by: $E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b)$ The focus of this ellipse is at $(ae, 0)$, and its latus rectum is the vertical line segment passing through the focus with endpoints at: $\left(ae, \frac{b^2}{a}\right) \quad \text{and} \quad \left(ae, -\frac{b^2}{a}\right)$

3. Condition of Common Latus Rectum: According to the problem, the line segment joining the points of intersection of $P$ and $E$ is their common latus rectum. Thus, the endpoints of the latus rectum of $P$ must coincide with the endpoints of the latus rectum of $E$: $k = ae$ $2k = \frac{b^2}{a}$

4. Relating Eccentricity $e$: Substituting $k = ae$ into $2k = \frac{b^2}{a}$, we get: $2ae = \frac{b^2}{a}$

   For an ellipse, we know that $b^2 = a^2(1 - e^2)$. Substituting this into the equation above: $2ae = \frac{a^2(1 - e^2)}{a}$ $2ae = a(1 - e^2)$

   Since $a > 0$, we can divide both sides by $a$: $2e = 1 - e^2 \implies e^2 + 2e - 1 = 0$

5. Solving for $e$: Using the quadratic formula to solve for $e$: $e = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$

   Since the eccentricity of an ellipse must satisfy $0 < e < 1$, we select the positive root: $e = \sqrt{2} - 1$

6. Calculating $e^2 + 2\sqrt{2}$: Squaring $e$: $e^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$

   Now, substituting $e^2$ into $e^2 + 2\sqrt{2}$: $e^2 + 2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$

The value of $e^2 + 2\sqrt{2}$ is $3$.