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

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$-\frac{20}{9}$

Vedclass · Answer

(A) For the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 4$ 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 focus is $(ae, 0) = (2 	imes \frac{3}{2}, 0) = (3, 0)$.The extremity of the latus rectum in the first quadrant is $(ae, \frac{b^2}{a}) = (3, \frac{5}{2})$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ 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 $\frac{3x}{4} - \frac{(5/2)y}{5} = 1$,which simplifies to $\frac{3x}{4} - \frac{y}{2} = 1$.For point $A$ on the $x$-axis,set $y=0$: $\frac{3x}{4} = 1 \implies x = \frac{4}{3}$. So $A = (\frac{4}{3}, 0)$ and $(OA)^2 = \frac{16}{9}$.For point $B$ on the $y$-axis,set $x=0$: $-\frac{y}{2} = 1 \implies y = -2$. So $B = (0, -2)$ and $(OB)^2 = 4$.Thus,$(OA)^2 - (OB)^2 = \frac{16}{9} - 4 = \frac{16 - 36}{9} = -\frac{20}{9}$.

The tangent drawn 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. If $O$ is the origin,then $(OA)^2-(OB)^2=$

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