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(A) Since $	riangle OAB$ is an equilateral triangle and $O$ is the vertex $(0,0)$,the axis of the parabola (the $x$-axis) bisects the angle $\angle A

Vedclass · Accepted Answer

$8a\sqrt{3}$ unit

Vedclass · Answer

(A) Since $	riangle OAB$ is an equilateral triangle and $O$ is the vertex $(0,0)$,the axis of the parabola (the $x$-axis) bisects the angle $\angle AOB$. Thus,the angle that $OA$ makes with the $x$-axis is $30^{\circ}$. The slope of $OA$ is $m = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$. The equation of line $OA$ is $y = \frac{1}{\sqrt{3}}x$. Substituting $y = \frac{x}{\sqrt{3}}$ into the parabola equation $y^2 = 4ax$: $\left(\frac{x}{\sqrt{3}}ight)^2 = 4ax \Rightarrow \frac{x^2}{3} = 4ax$. Since $x 
eq 0$ for point $A$,we have $x = 12a$. Then $y = \frac{12a}{\sqrt{3}} = 4\sqrt{3}a$. Point $A$ is $(12a, 4\sqrt{3}a)$. Since $AB$ is perpendicular to the $x$-axis,the length $AB = 2y_A = 2(4\sqrt{3}a) = 8\sqrt{3}a$. Since $	riangle OAB$ is equilateral,the side length is $8\sqrt{3}a$.

$\triangle OAB$ is an equilateral triangle inscribed in the parabola $y^2 = 4ax, a > 0$ with $O$ as the vertex. Then the length of the side of $\triangle OAB$ is

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