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.

Asked by Sakshi | 9 months ago |  101

1 Answer

Solution :-

Let us consider OAB be the equilateral triangle inscribed in parabola y2 = 4ax.

Let AB intersect the x – axis at point C.

Now let OC = k

From the equation of the given parabola, we have,

So, y2 = 4ak

y =\(  ±2\sqrt{ak}\)

The coordinates of points A and B are (k, \(  2\sqrt{ak}\)), and (k, \( -  2\sqrt{ak}\))

AB = CA + CB

\(  2\sqrt{ak}+  2\sqrt{ak}\)

\(  4\sqrt{ak}\)

Since, OAB is an equilateral triangle, OA2 = AB2.

Then,

k2 + \( (2\sqrt{ak})^2\) = \( (  4\sqrt{ak})^2\)

k2 + 4ak = 16ak

k2 = 12ak

k = 12a

Thus, AB = \(4\sqrt{ak}=4\sqrt{(a×12a)}\) 

= \( 4\sqrt{12}a^2\)

= \( 4\sqrt{(4a×3a)}\)

= \( 4(2)\sqrt{3a}\)

\( 8\sqrt{3}a\)

Hence, the side of the equilateral triangle inscribed in parabola y2 = 4ax is \( 8\sqrt{3}a\)

Answered by Aaryan | 9 months ago

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