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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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