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  • An equilateral triangle is inscribed in the parabola y^2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

8. An equilateral triangle is inscribed in the parabola y^2 = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

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Given an equilateral triangle inscribed in parabola with the equation, y^2 = 4 ax

The one coordinate of the triangle is A (0, 0).

Now, let the other two coordinates of the triangle be 

B(x,\sqrt{4ax}) and C(x,-\sqrt{4ax})

Now, since the triangle is equilateral,

BC=AB=CA

2\sqrt{4ax}=\sqrt{(x-0)^2+(\sqrt{4ax}-0)^2}

x^2=12ax

x=12a

The coordinates of the points of the equilateral triangle are:

(0,0),(12,\sqrt{4a\times 12a}),(12,-\sqrt{4a\times 12a})=(0,0),(12,4\sqrt{3}a)\:and\:(12,-4\sqrt{3}a)

So, the side of the triangle is 

2\sqrt{4ax}=2\times4\sqrt{3}a=8\sqrt{3}a

 

Posted by

Pankaj Sanodiya

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