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If one end of a diameter of the ellipse $4 x^2+y^2=16$ is $(\sqrt{3}, 2)$, then the other end is:
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Verified Answer
The correct answer is:
$(-\sqrt{3},-2)$
Correct option is \(\mathrm{C}\)
Let \(P(h, k)\)
Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are \((-\sqrt{3},-2)\)
as, centre \((0,0)\) so, \(\frac{h+\sqrt{3}}{2}=0, \frac{k+2}{2}=0\)
So \((h, k)=(-\sqrt{3},-2)\)
So, \((h, k) \equiv(-\sqrt{3},-2)\)
Let \(P(h, k)\)
Since every diameter of an ellipse passes through the centre and is bisected by it, therefore the coordinates of the other end are \((-\sqrt{3},-2)\)
as, centre \((0,0)\) so, \(\frac{h+\sqrt{3}}{2}=0, \frac{k+2}{2}=0\)
So \((h, k)=(-\sqrt{3},-2)\)
So, \((h, k) \equiv(-\sqrt{3},-2)\)
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