$\begin {array} {1 1} (A)\;3\sqrt 8 a & \quad (B)\;8\sqrt 3 a \\ (C)\;24a & \quad (D)\;None \end {array}$

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- General equation of a parabola which is open leftwards is $y^2=4ax$

Step 1 :

Let 0AB be the equilateral triangle inscribed in parabola $y^2=4ax$

AB intersects the x - axis at point C.

Let 0C = k

Substituting in the general equation for x we get,

$y^2=4ak$

$ \Rightarrow y = \pm 2\sqrt{ak}$

Step 2 :

The coordinates of 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}$

0AB is an equilateral triangle.

$0A^2=0B^2$ ( Hint : But $0C^2+AC^2=0A^2$ )

(i.e) : $k^2+ (2 \sqrt{ak} )^2=(4 \sqrt{ak} )^2$

$ \Rightarrow k^2+4ak = 16ak$

$ \Rightarrow k^2=12ak$

$ k = 12a$

Step 3 :

$ \therefore AB = 4\sqrt{ak}$

$ = 4\sqrt{a \times 12a}$

$ 4\sqrt{12a^2}$

$ = 8\sqrt{3a}$

Hence the side of the equilateral triangle inscribed in parabola $y^2=4ax$ is $8\sqrt 3a$ units.

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