The directrix of the hyperbola frac{x^2}{9} - | vedclass

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Question

(A) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

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$x = 9/\sqrt{13}$

Vedclass · Answer

(A) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{4} = 1$.Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we get $a^2 = 9$ and $b^2 = 4$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{9}} = \sqrt{\frac{13}{9}} = \frac{\sqrt{13}}{3}$.The equation of the directrix for a hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $x = \pm \frac{a}{e}$.Substituting the values,$x = \pm \frac{3}{\sqrt{13}/3} = \pm \frac{9}{\sqrt{13}}$.Thus,the directrix is $x = 9/\sqrt{13}$.

The directrix of the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$ is:

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