If the tangent drawn at the point P(3 sqrt{2} | vedclass

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(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.The equation of the ta

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$\frac{12 \sqrt{2}-20}{5}$

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(C) The equation of the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$. Here $a^2 = 9$ and $b^2 = 16$,so $a = 3$ and $b = 4$.The equation of the tangent at $P(x_1, y_1) = (3 \sqrt{2}, 4)$ is given by $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$.Substituting the values,we get $\frac{x(3 \sqrt{2})}{9} - \frac{y(4)}{16} = 1$,which simplifies to $\frac{x \sqrt{2}}{3} - \frac{y}{4} = 1$.The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$.The directrix of the hyperbola is $x = \frac{a}{e} = \frac{3}{5/3} = \frac{9}{5}$.Since the point $Q(\alpha, \beta)$ lies on the directrix,$\alpha = \frac{9}{5}$.Substituting $x = \frac{9}{5}$ into the tangent equation: $\frac{(9/5) \sqrt{2}}{3} - \frac{y}{4} = 1$.$\frac{3 \sqrt{2}}{5} - \frac{y}{4} = 1 \implies \frac{y}{4} = \frac{3 \sqrt{2}}{5} - 1 = \frac{3 \sqrt{2} - 5}{5}$.Thus,$y = \beta = \frac{12 \sqrt{2} - 20}{5}$.Since $Q$ is in the fourth quadrant,we check the sign. $\sqrt{2} \approx 1.414$,so $12(1.414) \approx 16.968$. $16.968 - 20 < 0$,which is consistent with the fourth quadrant.

If the tangent drawn at the point $P(3 \sqrt{2}, 4)$ on the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ meets its directrix at $Q(\alpha, \beta)$ in the fourth quadrant,then $\beta=$

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