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

Solution:

Let ΔOAB be the equilateral triangle inscribed in the parabola y² = 4ax .

Let AB intersect the x-axis at point C

Let OC = k

From the equation of the given parabola, we have

⇒ y² = 4ak

⇒ y = ± 2√ak

Therefore,

the respective coordinates of points A and B are (k, 2√ak) and (k, - 2√ak)

Hence,

AB = AC + CB

= 2√ak + 2√ak

= 4√ak

Since, ΔOAB is an equilateral triangle,

⇒ OA² = AB²

⇒ OC² + AC² = AB²

Therefore,

⇒ k² + (2√ak)² = (4√ak)²

⇒ k ² + 4ak = 16ak

⇒ k ² = 12ak

⇒ k = 12a

Hence,

AB = 4√ak = 4√a × 12a

= 4√12a²

= 8√3 a

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NCERT Solutions Class 11 Maths Chapter 11 Exercise ME Question 8

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

Summary:

The side of the equilateral triangle inscribed in the parabola y² = 4ax is 8√3 a