Let F1 and F2 be the foci of the hyperbola C: x2/a2 − y2/b2 = 1 (a > 0, b >…

Concept. For a hyperbola x2/a2 − y2/b2 = 1 the foci lie at (±c, 0) with c2 = a2 + b2, the eccentricity is e = c/a, and hence b2/a2 = e2 − 1. The point of such a curve lying directly above a focus has ordinate b2/a (the semi-latus rectum). And if H is the foot of the perpendicular dropped from a point P onto a line passing through a point Q, then |PH| = |PQ| · sin θ, where θ is the angle that line makes with PQ at Q.

Application. Translate each given condition into coordinates.

1. Place F1(−c, 0), F2(c, 0) and O(0, 0), where c = ae.

2. MF2 ⊥ F1F2 says MF2 is perpendicular to the X-axis, so M lies directly above F2; that is, the abscissa of M is c.

3. Putting x = c in x2/a2 − y2/b2 = 1 gives y2 = b2(c2/a2 − 1) = b4/a2, so (taking M above the X-axis) M = (c, b2/a).

4. H lies on MF1 with MF1 ⊥ OH, so H is the foot of the perpendicular from O onto the line MF1; therefore |OH| is the distance from O to that line.

5. Let θ = ∠MF1F2, the angle the line MF1 makes with the X-axis at F1. In the right triangle OHF1, |OH| = |OF1| · sin θ = c · sin θ.

6. Since |OF2| = c, the condition |OH| = λ|OF2| becomes c · sin θ = λc, i.e. λ = sin θ.

7. From the coordinates of F1 and M, tan θ = (b2/a) / (2c) = (c2 − a2)/(2ac) = (e2 − 1)/(2e). Write t = tan θ.

8. Solve t = (e2 − 1)/(2e) for e: e2 − 2te − 1 = 0, so e = t + √(t2 + 1), taking the positive root because e > 1.

9. λ = 2/5 ⇒ sin θ = 2/5 ⇒ t = tan θ = 2/√21; then √(t2 + 1) = √(4/21 + 1) = 5/√21, so e = 2/√21 + 5/√21 = 7/√21 = √(49/21) = √(7/3).

10. λ = 3/5 ⇒ sin θ = 3/5 ⇒ t = tan θ = 3/4; then √(t2 + 1) = √(9/16 + 1) = 5/4, so e = 3/4 + 5/4 = 2.

11. Each map λ ↦ t ↦ e is strictly increasing, so the open interval λ ∈ (2/5, 3/5) carries over to an open interval of e with those two endpoints.

Cross-check. Take λ = 1/2, which lies inside (2/5, 3/5). Then sin θ = 1/2, so t = tan θ = 1/√3 and e = 1/√3 + √(1/3 + 1) = 1/√3 + 2/√3 = √3 ≈ 1.732. That value sits strictly between √(7/3) ≈ 1.528 and 2, and it fails every other offered interval: (√2, √3) and (√3, 2) both exclude √3 as an endpoint, and (1, √(7/3)) stops below 1.528.

Result. e ∈ (√(7/3), 2).