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(A) The given equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Distance between the foci $= 2ae$. Distance between the direct

Vedclass · Accepted Answer

$\sqrt{2}: 1$

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(A) The given equation of the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. Distance between the foci $= 2ae$. Distance between the directrices $= \frac{2a}{e}$. Given the ratio is $3: 2$,so $\frac{2ae}{2a/e} = \frac{3}{2}$. This simplifies to $e^{2} = \frac{3}{2}$. We know that for a hyperbola,$e = \sqrt{1 + \frac{b^{2}}{a^{2}}}$,so $e^{2} = 1 + \frac{b^{2}}{a^{2}}$. Substituting $e^{2} = \frac{3}{2}$,we get $1 + \frac{b^{2}}{a^{2}} = \frac{3}{2}$. $\frac{b^{2}}{a^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$. Therefore,$\frac{b}{a} = \frac{1}{\sqrt{2}}$,which implies $\frac{a}{b} = \sqrt{2}: 1$.

If the distance between the foci and the distance between the directrices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are in the ratio $3: 2$,then $a: b$ is

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