If one end of a diameter of the ellipse ( 4x^2 + y^2 = 16) i... | Mathematics | SolveeIt | Solveeit

Today, August 18

Question

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

A

$(-\sqrt{3},2)$

B

$(\sqrt{3},-2)$

C

$(\sqrt{3},\sqrt{2})$

D

(0, 0)

Answer

$(\sqrt{3},\sqrt{2})$

Explanation

Solution

[Hint. Since every diameter of an ellipse passes through the centre and is bisected by it, .-. the other end is $( - \sqrt3 , -\sqrt2 )$ ]