The angle of intersection of the curves xy = a2 and x2 − y2 = 2a2 is ______________ .

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$90^\circ$ \text { Given }: xy = a^2 . . . \left( 1 \right) x^2 - y^2 = 2 a^2 . . . \left( 2 \right) \text { Let} \left( x_1 , y_1 \right)\text {be the point of intersection }. \text { On differentiating (1) w.r.t. x, we get } xdy/dx + y = 0 \Rightarrow dy/dx = - y/x \Rightarrow m_1 = \left( dy/dx \right)_\left( x_1 , y_1 \right) = - y_1/x_1 \text { On differentiating (2) w.r.t.x, we get } 2x - 2y dy/dx = 0 \Rightarrow dy/dx = x/y \Rightarrow m_2 = \left( dy/dx \right)_\left( x_1 , y_1 \right) = x_1/y_1 \text { Now,} m_1 $\times$ m_2 = - y_1/x_1 $\times$ x_1/y_1 \Rightarrow m_1 $\times$ m_2 = - 1 \because m_1 $\times$ m = - 1 \text { So, the angle between the curves is } $90^\circ$

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