An equilateral triangle is inscribed in the parabola `y^2=4ax` where are at the vertex of the parabola. find the length of the side of the triangle.

Explanation

Equation of parabola `y^(2)=4ax`
`:. "Vertex"O-=(0,0)`
Let OAB is an equilateral triangle whose vertices lie on the given parabola.
In `DeltaAMO,angleAOB=60^(@)`
`rArr" "angleAOM=30^(@)`
`sin30^(@)=(AM)/(OA)rArr(1)/(2)=(AM)/(OA)`
`rArr" "AM=(1)/(2)*OA`
`and" "cos30^(@)=(OM)/(OA)rArrsqrt(3)/(2)=(OM)/(OA)`
`rArr" "OM=(sqrt(3))/(2)OA`
Coordinates of point
`A-=(OM,AM)`
`-=((sqrt(3))/(2)OA,(1)/(2)OA)`
This point lies on the parabola `y^(2)=4ax`
`rArr" "OA=8asqrt(3)`
`:." "((1)/(2)OA)^(2)=4a((sqrt(3))/(2)OA)`
`rArr" "OA=8asqrt(3)`
`:.` Length of sides of equilateral triangle `=8asqrt(3)` units