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An equilateral triangle is inscribed in ...

An equilateral triangle is inscribed in the parabola `y^(2)=4ax`,
such that one vertex of this coincides with the vertex of the parabola. Then find the side the length of this triangle.

A

`2a(2-sqrt(3))`

B

`4a(2-sqrt(3))`

C

`a(2-sqrt(3))`

D

`8a(2-sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
B
(2) Let `A-=(at_(1)^(2),2at_(1)),B=-(at_(2)^(2),-2at_(1))`.
We have
`m_(AS)=tan((5pi)/(6))`
`or(2at_(1))/(at_(1)^(2)-a)=-(1)/(sqrt(3))`
`ort_(1)^(2)+2sqrt(3)t_(1)-1=0`
`ort_(1)=-sqrt(3)pm2`
Clearly, `t_(1)=-sqrt(3)-2` is rejected.
Thus, `t_(1)=(2-sqrt(3))`.
Hence, `AB=4at_(1)=4a(2-sqrt(3))`.
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