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(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.Here,$a^2 = 4$ and $b^2 = 5$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 +

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$-\frac{20}{9}$

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(A) Given the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$.Here,$a^2 = 4$ and $b^2 = 5$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{5}{4}} = \sqrt{\frac{9}{4}} = \frac{3}{2}$.The extremity of the latus rectum in the first quadrant is $L = (ae, \frac{b^2}{a}) = (2 	imes \frac{3}{2}, \frac{5}{2}) = (3, \frac{5}{2})$.The equation of the tangent at $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.Substituting $(x_1, y_1) = (3, \frac{5}{2})$,we get $\frac{3x}{4} - \frac{y(5/2)}{5} = 1$,which simplifies to $\frac{3x}{4} - \frac{y}{2} = 1$.This can be written as $\frac{x}{4/3} + \frac{y}{-2} = 1$.The $x$-intercept is $OA = \frac{4}{3}$ and the $y$-intercept is $OB = -2$.Therefore,$(OA)^2 - (OB)^2 = (\frac{4}{3})^2 - (-2)^2 = \frac{16}{9} - 4 = \frac{16 - 36}{9} = -\frac{20}{9}$.

The tangent at an extremity (in the first quadrant) of the latus rectum of the hyperbola $\frac{x^2}{4} - \frac{y^2}{5} = 1$ meets the $x$-axis and $y$-axis at $A$ and $B$ respectively. Then $(OA)^2 - (OB)^2$,where $O$ is the origin,equals

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