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An equilateral triangle is inscribed in ...

An equilateral triangle is inscribed in an ellipse whose equation is `x^2+4y^2=4` If one vertex of the triangle is (0,1) then the length of each side is

A

`(8sqrt(3))/(13)`

B

`(24sqrt(3))/(13)`

C

`(16sqrt(3))/(13)`

D

`(48sqrt(3))/(13)`

Text Solution

Verified by Experts

The correct Answer is:
C
We have `(x^(2))/(4)+y^(2)=1" "(1)`

Equation of the sides AB is
`y-1=sqrt(3)(x-0)`
`rArr y=sqrt(3)x+1 " "(2)`
Solving (1) and (2), we get
`(x^(2))/(4)+(sqrt3x+1)^(2)=1`
`rArrx^(2)+4(3x^(2)+1+2sqrt(3)x)=4`
`rArr13x^(2)+8sqrt(3)x=0`
`rArx=0 or x =-(8sqrt(3))/(13)`
Putting `x=-(8sqrt(3))/(13)` in (2), we get
`y=-sqrt(3)((8sqrt(3))/(13))+1=1-(24)/(13)=-(11)/(13)`
Hence, `B-=((8sqrt(3))/(13),(11)/(13))`
`:.C-=((8sqrt(3))/(13),(11)/(13))`
`:.BC-=(16sqrt(3))/(13)`
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