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(C) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2}

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$2$

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(C) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$.Since the tangent passes through $C(0, -b)$,we have $\frac{0 \cdot x_1}{a^2} - \frac{(-b)y_1}{b^2} = 1$,which simplifies to $\frac{y_1}{b} = 1$,so $y_1 = b$.Substituting $y_1 = b$ into the hyperbola equation: $\frac{x_1^2}{a^2} - \frac{b^2}{b^2} = 1$ $\Rightarrow \frac{x_1^2}{a^2} = 2$ $\Rightarrow x_1 = \pm a\sqrt{2}$.Thus,the points of contact are $A(a\sqrt{2}, b)$ and $B(-a\sqrt{2}, b)$.The triangle $\Delta ABC$ has vertices $C(0, -b)$,$A(a\sqrt{2}, b)$,and $B(-a\sqrt{2}, b)$.The length of the base $AB$ is $2a\sqrt{2}$. The height of the triangle from $C$ to $AB$ is $h = b - (-b) = 2b$.For $\Delta ABC$ to be a right-angled triangle,the angle at $C$ must be $90^{\circ}$ (since $AB$ is horizontal and $C$ lies on the $y$-axis,the triangle is isosceles with $CA=CB$).If $\angle ACB = 90^{\circ}$,then the slope of $CA$ is $m_1 = \frac{b - (-b)}{a\sqrt{2} - 0} = \frac{2b}{a\sqrt{2}} = \frac{b\sqrt{2}}{a}$.The slope of $CB$ is $m_2 = \frac{b - (-b)}{-a\sqrt{2} - 0} = -\frac{b\sqrt{2}}{a}$.Since $m_1 \cdot m_2 = -1$,we have $-(\frac{b\sqrt{2}}{a})^2 = -1$ $\Rightarrow \frac{2b^2}{a^2} = 1$ $\Rightarrow \frac{a^2}{b^2} = 2$.

Let tangents drawn from point $C(0,-b)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ touch the hyperbola at points $A$ and $B$. If $\Delta ABC$ is a right-angled triangle,then $\frac{a^2}{b^2}$ is equal to -

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