Question

The foci of the hyperbola \[ 4x^2 - 9y^2 - 1 = 0 \] are:

• A. \( (\pm \sqrt{13}, 0) \)
• B. \( \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \)
• C. \( \left( 0, \pm \frac{\sqrt{3}}{6} \right) \)
• D. None of these

Hint:

For hyperbolas, the formula for foci is \( (\pm ae, 0) \), and eccentricity is found using \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

Answer 1

The Correct Option is B

Given equation: \[ 4x^2 - 9y^2 - 1 = 0 \] Rearrange: \[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1 \] \[ \frac{x^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 \] Comparing with the standard form: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] we get: \[ a = \frac{1}{2}, \quad b = \frac{1}{3} \] The eccentricity of a hyperbola is: \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] \[ = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{4}}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3} \] Foci are given by: \[ (\pm ae, 0) = \left( \pm \frac{1}{2} \times \frac{\sqrt{13}}{3}, 0 \right) \] \[ = \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \] Thus, the foci are: \[ \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \]

Answer 2

The foci of the hyperbola
\[ 4x^2 - 9y^2 - 1 = 0 \] To find the foci:
1. Rearrange the equation of the hyperbola in standard form: \[ 4x^2 - 9y^2 = 1 \] Divide through by 1: \[ \frac{x^2}{\frac{1}{4}} - \frac{y^2}{\frac{1}{9}} = 1 \] This is now in the standard form of a hyperbola: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \(a^2 = \frac{1}{4}\) and \(b^2 = \frac{1}{9}\), so \(a = \frac{1}{2}\) and \(b = \frac{1}{3}\). 2. The distance of the foci from the center is given by \(c = \sqrt{a^2 + b^2}\): \[ c = \sqrt{\frac{1}{4} + \frac{1}{9}} = \sqrt{\frac{13}{36}} = \frac{\sqrt{13}}{6} \] 3. The foci of the hyperbola are at \(\left( \pm \frac{\sqrt{13}}{6}, 0 \right)\). Thus, the foci are:

\( \left( \pm \frac{\sqrt{13}}{6}, 0 \right) \)