If theta is the acute angle between the tange | vedclass

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(C) The given equation of the hyperbola is $5x^2 - 6y^2 = 30$,which can be written as $\frac{x^2}{6} - \frac{y^2}{5} = 1$.Here,$a^2 = 6$ and $b^2 = 5$

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$\frac{4}{3}$

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(C) The given equation of the hyperbola is $5x^2 - 6y^2 = 30$,which can be written as $\frac{x^2}{6} - \frac{y^2}{5} = 1$.Here,$a^2 = 6$ and $b^2 = 5$.Let the equation of the tangent be $y = mx + c$. Since it passes through $(2, 3)$,we have $3 = 2m + c$,so $c = 3 - 2m$.The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.Substituting the values,we get $(3 - 2m)^2 = 6m^2 - 5$.$9 - 12m + 4m^2 = 6m^2 - 5$.$2m^2 + 12m - 14 = 0$,which simplifies to $m^2 + 6m - 7 = 0$.Factoring gives $(m + 7)(m - 1) = 0$,so the slopes are $m_1 = 1$ and $m_2 = -7$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1m_2} ight|$.$	an 	heta = \left| \frac{1 - (-7)}{1 + (1)(-7)} ight| = \left| \frac{8}{1 - 7} ight| = \left| \frac{8}{-6} ight| = \frac{4}{3}$.

If $\theta$ is the acute angle between the tangents drawn from the point $(2,3)$ to the hyperbola $5x^2-6y^2-30=0$,then $\tan \theta=$

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