Question:
An equilateral triangle is inscribed in the parabola y$^2$ = 4x one of whose vertex is at the vertex of the parabola, the length of each side of the triangle is
An equilateral triangle is inscribed in the parabola y$^2$ = 4x one of whose vertex is at the vertex of the parabola, the length of each side of the triangle is
Updated On: Apr 30, 2026
- $\frac{\sqrt{3}}{2}$
- $4\frac{\sqrt{3}}{2}$
- $8\frac{\sqrt{3}}{2}$
- $8\sqrt{3}$
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The Correct Option is D
Solution and Explanation
The correct option is (D): \(8\sqrt{3}\).
Let \(AB = \ell\). Since the line makes an angle of \(30^\circ\) with the axes:
\(AM = \ell \cos 30^\circ = \frac{\ell \sqrt{3}}{2}\)
\(BM = \ell \sin 30^\circ = \frac{\ell}{2}\)
Hence, coordinates of \(B\) are:
\(\left( \frac{\ell \sqrt{3}}{2}, \frac{\ell}{2} \right)\)
Since \(B\) lies on the parabola \(y^2 = 4x\), substituting gives:
\(\left(\frac{\ell}{2}\right)^2 = 4\left(\frac{\ell \sqrt{3}}{2}\right)\)
\(\frac{\ell^2}{4} = 2\ell \sqrt{3}\)
Solving:
\(\ell = 8\sqrt{3}\)
Therefore, the required value is \(8\sqrt{3}\).
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.