If l_1 and l_2 are the lengths of the perpend | vedclass

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(C) The given hyperbola is $5x^2 - 4y^2 = 20$,which can be written as $\frac{x^2}{4} - \frac{y^2}{5} = 1$. Its asymptotes are given by $\frac{x^2}{4}

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$\frac{4}{81}$

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(C) The given hyperbola is $5x^2 - 4y^2 = 20$,which can be written as $\frac{x^2}{4} - \frac{y^2}{5} = 1$. Its asymptotes are given by $\frac{x^2}{4} - \frac{y^2}{5} = 0$,which implies $\sqrt{5}x - 2y = 0$ and $\sqrt{5}x + 2y = 0$. Let $(x_0, y_0)$ be a point on the hyperbola,so $5x_0^2 - 4y_0^2 = 20$. The lengths of the perpendiculars from $(x_0, y_0)$ to the asymptotes are $l_1 = \frac{|\sqrt{5}x_0 - 2y_0|}{\sqrt{(\sqrt{5})^2 + (-2)^2}} = \frac{|\sqrt{5}x_0 - 2y_0|}{3}$ and $l_2 = \frac{|\sqrt{5}x_0 + 2y_0|}{\sqrt{(\sqrt{5})^2 + 2^2}} = \frac{|\sqrt{5}x_0 + 2y_0|}{3}$. Thus,$l_1 l_2 = \frac{|5x_0^2 - 4y_0^2|}{9} = \frac{20}{9}$. Therefore,$\frac{l_1^2 l_2^2}{100} = \frac{(20/9)^2}{100} = \frac{400/81}{100} = \frac{4}{81}$.

If $l_1$ and $l_2$ are the lengths of the perpendiculars drawn from a point on the hyperbola $5x^2 - 4y^2 - 20 = 0$ to its asymptotes,then $\frac{l_1^2 l_2^2}{100} = $

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