Consider the parabola $$P: y^2 = 4kx$$ and the ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$. Let the line segment joining the points of intersection of $$P$$ and $$E$$, be their latus rectums. If the eccentricity of $$E$$ is $$e$$, then $$e^2 + 2\sqrt{2}$$ is equal to _____.

The parabola is $$P : y^{2}=4kx$$. For a parabola in standard form $$y^{2}=4kx$$:
  • Focus $$(k,0)$$
  • Axis along the $$x$$-axis
  • Latus-rectum is the line $$x=k$$ whose endpoints on the parabola are obtained by substituting $$x=k$$ into the equation:

$$y^{2}=4k(k)\;\Longrightarrow\;y=\pm 2k$$
Hence the latus-rectum endpoints of the parabola are $$\bigl(k,\,\pm 2k\bigr)$$ and its length is $$4k$$.

The ellipse is $$E : \dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ with major axis along the $$x$$-axis.
For such an ellipse (eccentricity $$e$$):
  • Foci $$(\pm ae,0)$$
  • Latus-rectum is the chord through a focus perpendicular to the major axis, i.e. the vertical line $$x=ae$$.
Substituting $$x=ae$$ in the ellipse equation gives

$$\dfrac{(ae)^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 \;\Longrightarrow\; e^{2}+\dfrac{y^{2}}{b^{2}}=1 \;\Longrightarrow\; y=\pm\dfrac{b^{2}}{a}$$

Thus the ellipse’s latus-rectum endpoints are $$\bigl(ae,\,\pm\dfrac{b^{2}}{a}\bigr)$$ and its length is $$\dfrac{2b^{2}}{a}$$.

According to the question, “the line segment joining the points of intersection of $$P$$ and $$E$$” is simultaneously the latus-rectum of each curve. Therefore the two endpoints just found must coincide:

$$k = ae \qquad\text{and}\qquad 2k = \dfrac{b^{2}}{a} \;-(1)$$

For an ellipse, the semi-minor axis satisfies $$b^{2}=a^{2}(1-e^{2})$$. Using this with equations $$(1)$$:

From $$k=ae$$, write $$k$$ in terms of $$a,e$$ and substitute into $$2k=\dfrac{b^{2}}{a}$$:

$$2(ae) = \dfrac{b^{2}}{a} \;\Longrightarrow\; 2a^{2}e = b^{2}$$

But $$b^{2}=a^{2}(1-e^{2})$$, hence

$$a^{2}(1-e^{2}) = 2a^{2}e \;\Longrightarrow\; 1-e^{2} = 2e \;\Longrightarrow\; e^{2} + 2e - 1 = 0$$

Solve the quadratic for $$e$$ (eccentricity of an ellipse lies between 0 and 1):

$$e = \dfrac{-2 \pm \sqrt{4+4}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$$

Taking the positive value, $$e = -1 + \sqrt{2}$$.

Now compute $$e^{2}+2\sqrt{2}$$:

$$e^{2} = (\sqrt{2}-1)^{2} = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$$

$$e^{2}+2\sqrt{2} = (3 - 2\sqrt{2}) + 2\sqrt{2} = 3$$

Therefore, the required value is 3.