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 28, 2024
- $\frac{\sqrt{3}}{2}$
- $4\frac{\sqrt{3}}{2}$
- $8\frac{\sqrt{3}}{2}$
- $8\sqrt{3}$
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The Correct Option is D
Solution and Explanation
The correct option is(D): \(8\sqrt{3}\).
Let AB = \(\ell , \, then \, \, AM \, = \ell cos 30^{?} \, = \frac{\ell \sqrt{3}}{2}\)
& BM = \(\ell \, sin 30^{?} \, = \frac{\ell}{2}\)
So, the coordinates of B are \(\bigg( \frac{\ell \sqrt{3}}{2}, \frac{\ell}{2}\bigg)\)
Since, B lies on \(y^2 \, = \, 4x\)
\(\therefore \, \, \frac{\ell^2}{4}=4\bigg(\frac{\ell \sqrt{3}}{2}\bigg)\)
\(\Rightarrow \, \, \ell^2 \, \frac{16}{2}. \sqrt{3\ell} \, \, \Rightarrow \, \, =8\sqrt{3}\)
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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.