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Let $A$, $B$ and $C$ are the vertices of an equilateral triangle and lies on the parabola $y^2=4ax$ where $A$ is the vertex of parabola. Find the side length of triangle.

My attempt :

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Slope of line $AB=\dfrac {2}{t}$ where $t$ is a parameter. So, $$\tan (30)=\dfrac {2}{t}$$ $$\dfrac {1}{\sqrt {3}}=\dfrac {2}{t}$$ $$t=2\sqrt {3}$$

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1)Equation of line $AB$:

$y=(√3/3)x$; where $m =\tan 30° =√3/3.$

2) Intersection with parabola $y^2=4ax:$

$y^2= 4a(3/√3)y;$

2 solutions: $y=0$, and

$y= 4a(3/√3)= (12a)/√3.$

Corresponding to $x=0$ and :

$x = (3/√3)((12a)/√3)= 12a.$

Length of a side of the equilateral $\triangle ABC$:

Length $BC:$ $(24a)/√3.$

Check: Length $AB:$

$\sqrt{(12a)/√3)^2 + (12a)^2}=$

$12a\sqrt{1/3+1}= (24a)/√3.$

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  • $\begingroup$ amd..A typo: should be BC, as you suggest.Thanks. $\endgroup$ – Peter Szilas Jun 2 '18 at 20:44

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