A hyperbola with centre at (0,0) has its tran | vedclass

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(B) The equation of the hyperbola with centre $(0,0)$ and transverse axis along the $X$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given that the

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$\frac{2 \sqrt{2}}{\sqrt{7}}$

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(B) The equation of the hyperbola with centre $(0,0)$ and transverse axis along the $X$-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Given that the length of the transverse axis is $2a = 12$,so $a = 6$.Substituting $a=6$ into the equation,we get $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$.Since the point $(8,2)$ lies on the hyperbola,we have $\frac{8^2}{36} - \frac{2^2}{b^2} = 1$.$\frac{64}{36} - \frac{4}{b^2} = 1 \implies \frac{16}{9} - 1 = \frac{4}{b^2}$.$\frac{7}{9} = \frac{4}{b^2} \implies b^2 = \frac{36}{7}$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$.$e = \sqrt{1 + \frac{36/7}{36}} = \sqrt{1 + \frac{1}{7}} = \sqrt{\frac{8}{7}} = \frac{2\sqrt{2}}{\sqrt{7}}$.

$A$ hyperbola with centre at $(0,0)$ has its transverse axis along the $X$-axis,and its length is $12$. If $(8,2)$ is a point on the hyperbola,then its eccentricity is

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