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Mathematics Question on Parabola

An equilateral triangle is inscribed in the parabola y2^2 = 4x one of whose vertex is at the vertex of the parabola, the length of each side of the triangle is

A

32\frac{\sqrt{3}}{2}

B

4324\frac{\sqrt{3}}{2}

C

8328\frac{\sqrt{3}}{2}

D

838\sqrt{3}

Answer

838\sqrt{3}

Explanation

Solution

The correct option is(D): 838\sqrt{3}.

Let AB = ℓ, then  AM =ℓcos30? =ℓ32\ell , \, then \, \, AM \, = \ell cos 30^{?} \, = \frac{\ell \sqrt{3}}{2}
& BM = ℓ sin30? =ℓ2\ell \, sin 30^{?} \, = \frac{\ell}{2}

So, the coordinates of B are (ℓ32,ℓ2)\bigg( \frac{\ell \sqrt{3}}{2}, \frac{\ell}{2}\bigg)
Since, B lies on y2 = 4xy^2 \, = \, 4x
∴  ℓ24=4(ℓ32)\therefore \, \, \frac{\ell^2}{4}=4\bigg(\frac{\ell \sqrt{3}}{2}\bigg)
⇒  ℓ2 162.3ℓ  ⇒  =83\Rightarrow \, \, \ell^2 \, \frac{16}{2}. \sqrt{3\ell} \, \, \Rightarrow \, \, =8\sqrt{3}