Given \(x^2 + 3y^2 = 9\), i.e., \(\dfrac{x^2}{9} + \dfrac{y^2}{3} = 1\). The equation of tangent at point \((3\cos\theta, \sqrt{3}\sin\theta)\) is \(\dfrac{x\cos\theta}{3} + \dfrac{y\sin\theta}{\sqrt{3}} = 1\). The equation of tangent at point \((-3\sin\theta, \sqrt{3}\cos\theta)\) is \(\dfrac{-x\sin\theta}{3} + \dfrac{y\cos\theta}{\sqrt{3}} = 1\). For the two tangents to be perpendicular, which of the following is correct?
Let the ellipse E : xa 2 + yb 2 = 1, a > b and E 2 : xA 2 + yB 2 = 1, A < B have same eccentricity 1 1 32 . Let the product of their lengths of latus rectums be , and the distance between the foci of E 1 be 4. If 3 3 E 1 and E 2 meet at A, B, C and D , then the area of the quadrilateral ABCD equals : 6