Question

The length of the side of an equilateral triangle inscribed in the parabola, $y^2=4x$ so that one of its angular point is at the vertex is

• A. $8\sqrt{3}$
• B. $6\sqrt{3}$
• C. $4\sqrt{3}$
• D. $2\sqrt{3}$

Answer 1

Answer: (A)

$8\sqrt{3}$

The length of side of equilateral triangle inscribed in a parabola, $y^2=4x$

So that one of its angular point is at the vertex is

Let OAB is equilateral triangle

$OA=AB=AB=x$

OC is perpendicular to AB

Angle OCA is equal to angle OCB

$OC=OCOA=OB△OAC\cong △OBCAC=BC=\frac{x}{2}△AOC(OA{)}^2=(OC{)}^2+(AC{)}^2{x}^2=(OC{)}^2+\frac{x^2}{4}OC=\frac{\sqrt{3}}{2}xA=(\frac{\sqrt{3}}{2}x,\frac{x}{2})$

Point A lies on parabola $y^2=4x$

${\left(\frac{x}{2}\right)}^2=4\left(\frac{\sqrt{3}}{2}x\right)x^2-8\sqrt{3}x=0x=0,8\sqrt{3}$

$0$ is not accepted

Therefore $x=8\sqrt{3}$