If a circle of radius 4 text{ cm} passes thro | vedclass

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(A) The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$. Here,$b^2 = 4$,so $b = 2$.Since the circle is concentric with the hyperbola and pass

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(A) The given hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$. Here,$b^2 = 4$,so $b = 2$.Since the circle is concentric with the hyperbola and passes through its foci $(\pm c, 0)$,the radius of the circle is equal to the distance of the foci from the center $(0, 0)$.Thus,$c = 4$.Using the relation $c^2 = a^2 + b^2$,we have $16 = a^2 + 4$,which gives $a^2 = 12$,so $a = 2 \sqrt{3}$.The conjugate hyperbola is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$,which is $\frac{y^2}{4} - \frac{x^2}{12} = 1$.For this conjugate hyperbola,the semi-transverse axis is $b = 2$ and the semi-conjugate axis is $a = 2 \sqrt{3}$.Let $e'$ be the eccentricity of the conjugate hyperbola. The relation is $a^2 = b^2(e'^2 - 1)$ for the conjugate hyperbola,or more simply,$c^2 = a^2 + b^2$ where $c$ is the distance to the foci of the conjugate hyperbola.For the conjugate hyperbola,the foci are on the $y$-axis at $(0, \pm c')$,where $c'^2 = a^2 + b^2 = 12 + 4 = 16$,so $c' = 4$.The eccentricity $e'$ is given by $c' = b e'$,where $b$ is the semi-transverse axis of the conjugate hyperbola.Thus,$4 = 2 e'$,which gives $e' = 2$.

If a circle of radius $4 \text{ cm}$ passes through the foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$ and is concentric with the hyperbola,then the eccentricity of the conjugate hyperbola of that hyperbola is

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