Question

The descending order of magnitude of the eccentricities of the following hyperbolas is: A. A hyperbola whose distance between foci is three times the distance between its directrices. B. Hyperbola in which the transverse axis is twice the conjugate axis. C. Hyperbola with asymptotes \( x + y + 1 = 0, x - y + 3 = 0 \).

• A. \( C, A, B \)
• B. \( B, C, A \)
• C. \( \) (No option provided)
• D. \( A, C, B \)

Hint:

The eccentricity of a hyperbola is always greater than 1 and can be determined using the transverse and conjugate axis relations.

Answer 1

The Correct Option is D

Step 1: Finding the eccentricities For a hyperbola, the eccentricity is given by: \[ e = \frac{\text{distance between foci}}{\text{length of transverse axis}}. \] Solving for each case: - Hyperbola A: Given condition leads to \( e = \frac{3}{2} \). - Hyperbola B: \( e = \sqrt{5}/2 \). - Hyperbola C: Given asymptotes suggest \( e = \sqrt{2} \). Ordering the values, we get: \[ A > C > B. \]

Answer 2

Given three hyperbolas:

A. A hyperbola whose distance between foci is three times the distance between its directrices.
B. A hyperbola in which the transverse axis is twice the conjugate axis.
C. A hyperbola with asymptotes \( x + y + 1 = 0 \) and \( x - y + 3 = 0 \).

We need to arrange them in descending order of their eccentricities.

Step 1: Recall eccentricity \( e \) of a hyperbola is related to parameters:
For hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), eccentricity:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \] and transverse axis length \( 2a \), conjugate axis length \( 2b \).

Step 2: Hyperbola A:
Distance between foci = \( 2ae \).
Distance between directrices = \( \frac{2a}{e} \).
Given:
\[ 2ae = 3 \times \frac{2a}{e} \implies 2ae = \frac{6a}{e} \] Divide both sides by \( 2a \):
\[ e = \frac{3}{e} \implies e^2 = 3 \implies e = \sqrt{3} \]

Step 3: Hyperbola B:
Transverse axis = \( 2a \), conjugate axis = \( 2b \), with given ratio:
\[ 2a = 2 \times (2b) \implies a = 2b \] Calculate eccentricity:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{b^2}{(2b)^2}} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} \approx 1.118 \]

Step 4: Hyperbola C:
Asymptotes are:
\[ x + y + 1 = 0 \quad \Rightarrow \quad y = -x - 1 \] \[ x - y + 3 = 0 \quad \Rightarrow \quad y = x + 3 \] Slopes of asymptotes are \( m_1 = -1 \) and \( m_2 = 1 \).
For hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), asymptotes are:
\[ y = \pm \frac{b}{a} x \] So:
\[ \frac{b}{a} = 1 \implies b = a \] Calculate eccentricity:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2} \approx 1.414 \]

Step 5: Arrange eccentricities in descending order:
\[ e_A = \sqrt{3} \approx 1.732 \gt e_C = \sqrt{2} \approx 1.414 \gt e_B = \frac{\sqrt{5}}{2} \approx 1.118 \]

Final order:
\[ \boxed{A, C, B} \]