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Question: If one end of a diameter of the ellipse \(4x^{2} + y^{2} = 16\)is \((\sqrt{3},2)\), then the other e...

If one end of a diameter of the ellipse 4x2+y2=164x^{2} + y^{2} = 16is (3,2)(\sqrt{3},2), then the other end is

A

(−3,2)( - \sqrt{3},2)

B

(3,−2)(\sqrt{3}, - 2)

C

(−3,−2)( - \sqrt{3}, - 2)

D

(0,0)(0,0)

Answer

(−3,−2)( - \sqrt{3}, - 2)

Explanation

Solution

Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of theother end are (−3,−2)( - \sqrt{3}, - 2)