Question

The distance between the directrices of the hyperbola \( x^2 - y^2 = 9 \) is

• A. \( \frac{9}{\sqrt{2}} \)
• B. \( \frac{5}{\sqrt{3}} \)
• C. \( \frac{3}{\sqrt{2}} \)
• D. \( 3\sqrt{2} \)
• E. \( 5\sqrt{3} \)

Hint:

For hyperbola \( x^2/a^2 - y^2/b^2 = 1 \), always remember: Directrices = \( x = \pm \frac{a}{e} \) and distance = \( \frac{2a}{e} \).

Answer 1

The Correct Option is D

Concept:
• Standard equation of hyperbola: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
• Eccentricity: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
• Equations of directrices: \[ x = \pm \frac{a}{e} \]
• Distance between two parallel lines \( x = c \) and \( x = -c \) is \( 2c \)

Step 1: Convert given equation into standard form.
Given: \[ x^2 - y^2 = 9 \] Divide both sides by 9: \[ \frac{x^2}{9} - \frac{y^2}{9} = 1 \] Comparing with standard form: \[ a^2 = 9, \quad b^2 = 9 \] Thus, \[ a = 3, \quad b = 3 \]

Step 2: Find eccentricity of hyperbola.
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \] Substitute values: \[ e = \sqrt{1 + \frac{9}{9}} = \sqrt{1 + 1} = \sqrt{2} \]

Step 3: Write equations of directrices.
For hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), directrices are: \[ x = \pm \frac{a}{e} \] Substitute: \[ x = \pm \frac{3}{\sqrt{2}} \]

Step 4: Find distance between the two directrices.
The two lines are: \[ x = \frac{3}{\sqrt{2}} \quad \text{and} \quad x = -\frac{3}{\sqrt{2}} \] Distance between them: \[ \left| \frac{3}{\sqrt{2}} - \left(-\frac{3}{\sqrt{2}}\right) \right| \] \[ = \frac{6}{\sqrt{2}} \]

Step 5: Simplify the expression.
\[ \frac{6}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \]

Step 6: Final Answer.
\[ \boxed{3\sqrt{2}} \]