Question

If M is the foot of the perpendicular from a point P a parabola $y^2$ = 4ax to its directrix and SPM is an equilateral triangle, where S is the focus, then SP is equal to :

• A. $a$
• B. $2a$
• C. $3a$
• D. $4a$

Answer 1

The correct option is D $4a$

If $M$ is foot of perpendicular from a point $P$ a parabola $y^2=4ax$ to its directrix and $SPM$ is an equilateral triangle where $S$ is focus,then $SP$ is equal to,

Let

$=\left|\frac{3\left(2\right)-4\left(3\right)+7}{5}\right|=\frac{2}{5}{y}^2=4axP\equiv (a{t}^2,2at)SP=PM=MS\left(equilatral△\right)\sqrt{(a-(-a){)}^2+(2at-0{)}^2}=\sqrt{(a{t}^2-a{)}^2+(2at-0{)}^2}(2a{)}^2+(2at{)}^2=(a{t}^2-a{)}^2+(2at{)}^{2}4a^2=a^2{t}^4+a^2-2a^2{t}^2{a}^2{t}^4-2a^2{t}^2-3a^2=0t^4-2t^2-3=0t^2=\frac{2\pm \sqrt{4\pm 12}}{2}=\frac{2\pm 4}{2}\frac{6}{2},\frac{-2}{2}(3,-1)$

Since negative value is not accepted.

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