Let P, Q, R, S be the points of intersection | vedclass

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(B) Given the equations: $x^2+y^2=4$ $xy=\sqrt{3}$ Substitute $y=\frac{\sqrt{3}}{x}$ into the circle equation: $x^2 + \frac{3}{x^2} = 4$ $x^4 - 4x^2 +

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$x+\sqrt{3}y=2\sqrt{3}$

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(B) Given the equations: $x^2+y^2=4$ $xy=\sqrt{3}$ Substitute $y=\frac{\sqrt{3}}{x}$ into the circle equation: $x^2 + \frac{3}{x^2} = 4$ $x^4 - 4x^2 + 3 = 0$ $(x^2-3)(x^2-1) = 0$ So,$x^2=3$ or $x^2=1$. Since $\alpha > \beta > 0$,we have $x^2=3$ and $y^2=1$. Thus,$\alpha = \sqrt{3}$ and $\beta = 1$. Point $P = (\sqrt{3}, 1)$. The equation of the tangent to the hyperbola $xy=c^2$ at $(x_1, y_1)$ is $xy_1 + yx_1 = 2c^2$. Here $c^2 = \sqrt{3}$,$x_1 = \sqrt{3}$,$y_1 = 1$. So,$x(1) + y(\sqrt{3}) = 2\sqrt{3}$. $x + \sqrt{3}y = 2\sqrt{3}$.

Let $P, Q, R, S$ be the points of intersection of the circle $x^2+y^2=4$ and the hyperbola $xy=\sqrt{3}$. If $P=(\alpha, \beta)$ and $\alpha>\beta>0$,then the equation of the tangent drawn at $P$ to the hyperbola is

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