Misc 8
An equilateral triangle is inscribed in the parabola y2 = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
Let length of equilateral triangle be s
Hence OA = OB = AB = s
Here, OC AB
So, OCA = OCB = 90
And AC = BC
So, AC = BC = 2
AC = BC = 2
We find coordinates of point B
Now,
in right triangle OBC
Using Pythagoras theorem
(Hypotenuse)2 = (Height)2 + (Base)2
(OB)2 = (OC)2 + (BC)2
s2 = (OC)2 + 2 2
s2 = (OC)2 + 2 4
s2 2 4 = (OC)2
4 2 2 4 = (OC)2
3 2 4 = (OC)2
(OC)2 = 3 2 4
OC = 3 2 4
OC = 3 2
Hence coordinate of point B is B( , )
Now, point B lies on parabola
So, it must satisfy its equation
Putting x = 3 2 , y = 2 in equation of parabola
y2 = 4ax
2 2 = 4a( 3 2 )
2 4 = 4a( 3 2 )
2 = 4 4a( 3 2 )
s = 8 a
Hence, side of equilateral triangle = 8 3 a

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.