Let $F_{1}$ , $F_{2}$ be the foci of the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , $a > 0$ , $b > 0$ , and let O be the origin. Let M be a point on the hyperbola above the x-axis and H be a point on $M F_{1}$ such that $M F_{2} ⊥ F_{1} F_{2}$ , $M F_{1} ⊥ O H$ and $∣ O H ∣ = λ ∣ O F_{2} ∣$ with $λ \in (\frac{2}{5}, \frac{3}{5})$ . Then the range of the eccentricity $e$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $(\frac{7}{3}, 2)$

Detailed Solution

$M F_{2} ⊥ F_{1} F_{2}$ puts M at $(c, \frac{b ^{2}}{a})$ , so $M F_{2} = \frac{b ^{2}}{a}$ and, since $M F_{1} - M F_{2} = 2 a$ , $M F_{1} = 2 a + \frac{b ^{2}}{a}$ .

OH is the distance from O to $M F_{1}$ : $∣ O H ∣ = O F_{1} s in ∠ M F_{1} F_{2} = c \cdot \frac{M F _{2}}{M F _{1}} = c \cdot \frac{b ^{2}}{2 a ^{2} + b ^{2}}$ .

$λ = \frac{b ^{2}}{2 a ^{2} + b ^{2}} = \frac{e ^{2} - 1}{e ^{2} + 1}$ (using $b^{2} = a^{2} (e^{2} - 1)$ ).

$\frac{2}{5} < \frac{e ^{2} - 1}{e ^{2} + 1} < \frac{3}{5}$ gives $\frac{7}{3} < e^{2} < 4$ , so $\frac{7}{3} < e < 2$ .

✓ Check

e = 2

gives

λ = \frac{1}{3} < \frac{2}{5}

, so

2

lies outside the range, as (A) says.

⚠ Common trap

(C)

(1, \frac{7}{3})

is the range for

λ < \frac{2}{5}

: the first inequality reversed.

Quick tip

$λ = \frac{e ^{2} - 1}{e ^{2} + 1} \in (\frac{2}{5}, \frac{3}{5}) \Rightarrow \frac{7}{3} < e^{2} < 4$

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Hyperbola — Preparation

This question tests your understanding of Hyperbola, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A hyperbola is the set of points where the difference of the distances from two fixed points (the foci) is constant. Its standard form is x^2/a^2 - y^2/b^2 = 1, with eccentricity given by b^2 = a^2(e^2 - 1), where e > 1. The foci are at (ae, 0) and (-ae, 0), and the asymptotes are the lines y = (b/a)x and y = -(b/a)x. Exam questions test finding the foci, eccentricity, transverse and conjugate axes, latus rectum, asymptotes, and equations of tangents. A special case is the rectangular hyperbola, where a = b and e = sqrt 2. Many students confuse the hyperbola with the ellipse, so compare the signs in the equation and remember that the difference of focal distances equals 2a. Practise the hyperbola questions below, each with a step-by-step solution.

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