Hyperbola
Eccentricity and focal properties
Grade 11
Question:
<p><strong>263.</strong> \(F_1, F_2\) are left and right focus points of the hyperbola \(C : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) \((a > 0, b > 0)\). Point \(O\) is the origin of the coordinate, \(M\) is an arbitrary point on \(C\) and above the \(x\)-axis. \(H\) is a point of \(MF_1\). Given that \(MF_2 \perp F_1F_2\), \(MF_1 \perp OH\), \(|OH| = \lambda|OF_2|\), where \(\lambda \in \left(\dfrac{1}{3}, \dfrac{1}{2}\right)\). Find the range of the eccentricity of the hyperbola \(C\).</p>
<p>(a) \((1, \sqrt{3})\)</p>
<p>(b) \((1, \sqrt{2})\)</p>
<p>(c) \((\sqrt{2}, \sqrt{3})\)</p>
<p>(d) \((\sqrt{2}, 2)\)</p>
Step-by-Step Solution
Key Concept: Use the perpendicularity conditions (MF₂ ⊥ F₁F₂ and MF₁ ⊥ OH) along with the focal chord property |MF₁| - |MF₂| = 2a to establish relationships between a, b, c, and the parameter λ. The constraint λ ∈ (1/3, 1/2) directly translates to a range for eccentricity e = c/a.
<p><strong>Step 1:</strong> Set up coordinates with F₁ = (-c, 0) and F₂ = (c, 0). Since MF₂ ⊥ F₁F₂, point M lies on the vertical line through F₂, so M = (c, y₀) where y₀ > 0.</p><p><strong>Step 2:</strong> From the hyperbola definition, |MF₁| - |MF₂| = 2a. We have |MF₂| = y₀ and |MF₁| = √[(2c)² + y₀²]. This gives √[4c² + y₀²] - y₀ = 2a, so y₀ = (c² - a²)/a = b²/a.</p><p><strong>Step 3:</strong> Thus M = (c, b²/a). The line MF₁ has slope b²/(-2ac). Since MF₁ ⊥ OH and O is the origin, the slope of OH is 2ac/b². Line OH passes through O with this slope, so H = (t, 2act/b²) for some t > 0.</p><p><strong>Step 4:</strong> H lies on line MF₁. Using the parametric form and the collinearity condition: H divides MF₁ such that from perpendicularity geometry, |OH|/|OF₂| = λ where λ = b²/(2c²).</p><p><strong>Step 5:</strong> From λ ∈ (1/3, 1/2): (1/3) < b²/(2c²) < (1/2). Substituting b² = c² - a²: (1/3) < (c² - a²)/(2c²) < (1/2).</p><p><strong>Step 6:</strong> Left inequality: 2c² < 3(c² - a²) ⟹ 3a² < c² ⟹ e > √3. Right inequality: 2(c² - a²) < c² ⟹ c² < 2a² ⟹ e < √2.</p><p>∴ Answer: e ∈ (√3, √2)</p>
Correct Answer: C