If the circle x^2+y^2=a^2 intersects the hype | vedclass

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(C) We have the equations:$x^2+y^2=a^2$ $\dots(i)$$xy=b^2$ $\dots(ii)$From equation $(ii)$,we have $x = \frac{b^2}{y}$.Substituting this into equation

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$b^4$

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(C) We have the equations:$x^2+y^2=a^2$ $\dots(i)$$xy=b^2$ $\dots(ii)$From equation $(ii)$,we have $x = \frac{b^2}{y}$.Substituting this into equation $(i)$:$\left(\frac{b^2}{y}\right)^2 + y^2 = a^2$$\frac{b^4}{y^2} + y^2 = a^2$Multiplying by $y^2$:$b^4 + y^4 = a^2 y^2$$y^4 - a^2 y^2 + b^4 = 0$This is a biquadratic equation in $y$. Let its roots be $y_1, y_2, y_3, y_4$.According to the properties of the roots of a polynomial equation,the product of the roots $y_1 y_2 y_3 y_4$ is equal to the constant term divided by the leading coefficient.Thus,$y_1 y_2 y_3 y_4 = \frac{b^4}{1} = b^4$.

If the circle $x^2+y^2=a^2$ intersects the hyperbola $xy=b^2$ at four points $(x_1, y_1)$,$(x_2, y_2)$,$(x_3, y_3)$,and $(x_4, y_4)$,then $y_1 y_2 y_3 y_4 = $

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