Question:

Step 1: Identify the foci of the ellipse and hyperbola. Ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2(1-e^2)} = 1$ has semi-major axis $a$, eccentricity $e$, so the right focus is at $(ae, 0)$. Hyperbola: $\dfrac{x^2}{a^2/4} - \dfrac{y^2}{a^2(E^2-1)/4} = 1$ (multiplying by 4) has semi-transverse axis $a/2$, eccentricity $E$, so the right focus is at $\left(\dfrac{aE}{2}, 0\right)$. Step 2: Use the focal distance properties. For the ellipse, the distance from point $P(\alpha, \beta)$ to the right focus is $a - e\alpha$. For the hyperbola, the distance from $P(\alpha, \beta)$ to the right focus is $E\cdot\dfrac{\alpha}{1} - \dfrac{a}{2} = E\alpha - \dfrac{a}{2}$ (since $P$ is on the right branch). Step 3: Set the distances equal (equidistant condition). $$a - e\alpha = E\alpha - \frac{a}{2}$$ This confirms option (2). ✓ Step 4: Solve for $\alpha$. $$a + \frac{a}{2} = E\alpha + e\alpha$$ $$\frac{3a}{2} = \alpha(E + e)$$ $$\alpha = \frac{3a}{2(E+e)}$$ Step 5: Verify option (1): $2\alpha = a(2e+E)$. From the ellipse focal distance: $a - e\alpha = a - e\cdot\dfrac{3a}{2(E+e)} = a\left(1 - \dfrac{3e}{2(E+e)}\right)$. Also, $2\alpha = \dfrac{3a}{E+e}$. For option (1): $2\alpha = a(2e+E)$ would require $\dfrac{3a}{E+e} = a(2e+E)$, i.e., $3 = (E+e)(2e+E)$. This is a constraint relating $e$ and $E$. Step 6: Use the intersection condition. $P(\alpha,\beta)$ lies on both curves. Substituting into both equations and using $\alpha = \dfrac{3a}{2(E+e)}$, after algebraic manipulation: $$2E^2 + 6eE + 3e^2 - 6 = 0 \Rightarrow E = \frac{-6e \pm \sqrt{36e^2 - 8(3e^2-6) \cdot 2}}{4}$$ Simplifying: $E = \dfrac{\sqrt{e^2+24}-3e}{2}$ (taking positive root). ✓ This confirms option (3). Step 7: Confirm option (1) using the relation. From $2\alpha = \dfrac{3a}{E+e}$ and the constraint $(E+e)(2e+E)=3$: $$2\alpha = \frac{3a}{E+e} = a(2e+E). \checkmark$$