Question

A point on a parabola $y^2=4ax$ the foot of the perpendicular from it upon the directrix, and the focus are the vertices of an equilateral triangle. The focal distance of the point is equal to -

• A. $1/2$
• B. $a$
• C. $2a$
• D. $4a$

Answer 1

Answer: (D)

$4a$

A point on a parabola $y^2=4ax$ the foot of perpendicular from it upon the directrix and the focus are the vertices of an equilateral triangle

The focal distance of the point is equal to

APS is equilateral triangle

$AP=PS=SA$

$(a{t}^2-a{)}^2+(2at{)}^2=(2a{)}^2+(2at{)}^2{a}^2(t^2-1{)}^2=4a^2(t^2-1{)}^2=4t=\pm \sqrt{3}P\equiv (3a,\pm 2\sqrt{3}a)Q\equiv (a,0)PS=\sqrt{(3a-a{)}^2+(\pm 2\sqrt{3}a-0{)}^2}=\sqrt{4a^2+12a^2}=4a$