The point P(-2 sqrt{6}, sqrt{3}) lies on the | vedclass

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(C) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and eccentricity $e = \frac{\sqrt{5}}{2}$.We know $b^{2} = a^{2}(e^{2}-1) = a^{2}(

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$6 \sqrt{3}$

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(C) Given the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and eccentricity $e = \frac{\sqrt{5}}{2}$.We know $b^{2} = a^{2}(e^{2}-1) = a^{2}(\frac{5}{4}-1) = \frac{a^{2}}{4}$,so $a^{2} = 4b^{2}$.Since $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola,$\frac{(-2 \sqrt{6})^{2}}{4b^{2}} - \frac{(\sqrt{3})^{2}}{b^{2}} = 1$.$\frac{24}{4b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{6}{b^{2}} - \frac{3}{b^{2}} = 1 \Rightarrow \frac{3}{b^{2}} = 1 \Rightarrow b^{2} = 3$,so $b = \sqrt{3}$ and $a^{2} = 12$,so $a = 2 \sqrt{3}$.The hyperbola is $\frac{x^{2}}{12} - \frac{y^{2}}{3} = 1$.The tangent at $P(x_{1}, y_{1})$ is $\frac{xx_{1}}{a^{2}} - \frac{yy_{1}}{b^{2}} = 1$.$\frac{x(-2 \sqrt{6})}{12} - \frac{y(\sqrt{3})}{3} = 1 \Rightarrow -\frac{x \sqrt{6}}{6} - \frac{y}{\sqrt{3}} = 1$.For the conjugate axis,set $x = 0$: $-\frac{y}{\sqrt{3}} = 1 \Rightarrow y = -\sqrt{3}$. Thus $Q = (0, -\sqrt{3})$.The slope of the tangent is $m = \frac{b^{2}x_{1}}{a^{2}y_{1}} = \frac{3(-2 \sqrt{6})}{12(\sqrt{3})} = \frac{-6 \sqrt{6}}{12 \sqrt{3}} = -\frac{\sqrt{2}}{2} = -\frac{1}{\sqrt{2}}$.The slope of the normal is $m' = -\frac{1}{m} = \sqrt{2}$.The equation of the normal at $P$ is $y - \sqrt{3} = \sqrt{2}(x + 2 \sqrt{6})$.For the conjugate axis,set $x = 0$: $y - \sqrt{3} = \sqrt{2}(2 \sqrt{6}) = 2 \sqrt{12} = 4 \sqrt{3}$.$y = 4 \sqrt{3} + \sqrt{3} = 5 \sqrt{3}$. Thus $R = (0, 5 \sqrt{3})$.The distance $QR = |5 \sqrt{3} - (-\sqrt{3})| = |6 \sqrt{3}| = 6 \sqrt{3}$.

The point $P(-2 \sqrt{6}, \sqrt{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having eccentricity $e = \frac{\sqrt{5}}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively,then $QR$ is equal to :

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