Statement-I:Feet of perpendiculars drawn from foci of an ellipse $4 x^{2} + y^{2} = 16$ on the line $2 \sqrt{3} x + y = 8$ lie on the circle $x^{2} + y^{2} = 16$ .
Statement-II: If perpendiculars are drawn from foci of an ellipse to its tangent, the feet of these perpendiculars lie on director circle of the ellipse.

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Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is Ture; Statement-2 is Not a correct explanation for Statement-1

Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

answer is C.

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Detailed Solution

Simultaneously solving the equations of ellipse and the given line, we get
$4 x^{2} + {(8 - 2 \sqrt{3} x)}^{2} = 16 \Rightarrow x^{2} + {(4 - \sqrt{3} x)}^{2} = 4 \Rightarrow 4 x^{2} + 8 \sqrt{3} x + 12 = 0 \Rightarrow x^{2} - 2 \sqrt{3} x + 3 = 0 \Rightarrow {(x - \sqrt{3})}^{2} = 0$