Aspire Faculty ID #11967 · Topic: NIMCET 2025 · Just now
NIMCET 2025

An equilateral triangle is inscribed in the parabola $y^2 = x$. One vertex of the triangle is at the vertex of the parabola. The centroid of triangle is

Solution

Given parabola  $y^{2} = x. $ $ A(0,0) $ is one vertex of the equilateral triangle. 
Let the other two vertices be $ B(t^{2}, t) $ and $ C(t^{2}, -t) $, since they lie on the parabola and are symmetric. 
Distance: $ BC = \sqrt{(t^{2}-t^{2})^{2} + (t - (-t))^{2}} = 2t. $ 
And $ AB = \sqrt{(t^{2}-0)^{2} + (t-0)^{2}} = \sqrt{t^{4}+t^{2}} = t\sqrt{t^{2}+1}. $ 

For equilateral triangle: $ AB = BC $ 
$ t\sqrt{t^{2}+1} = 2t $ $ \sqrt{t^{2}+1} = 2 $ 
$ t^{2}+1 = 4 $ 
$ t^{2} = 3. $ 
So the points become: 
$ B = (3,\sqrt{3}), \quad C = (3,-\sqrt{3}). $ 
Centroid: $ G = \left( \frac{0+3+3}{3},; \frac{0+\sqrt3-\sqrt3}{3} \right) = (2,0). $
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