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

##### 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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