Question

Three normal are drawn from point $(5,0)$ to parabola $y^2=4x$. The centroid of the triangle formed by feet of these three normals is

• A. $(-3,\frac{2}{\sqrt{3}})$
• B. $(5,0)$
• C. $(2,0)$
• D. $(0,2)$

Answer 1

Answer: (A)

$(-3,\frac{2}{\sqrt{3}})$

Let the equation of any normal

$y=-tx+2t+t^3$

$(5,0)$ Passes through Normal

$0=-5t+2t+t^3$

$t(t^2-3t)=0$

$t=0$ $t=\pm \sqrt{3}$

$y=0$ $y=-\sqrt{3}x+2\sqrt{3}-3\sqrt{3}$

$x=-1,y=0$ $y=\sqrt{3}(-x+1)$

$x=-5y=0$ $y=\sqrt{3}(x+5)$

$x=-3y=2\sqrt{3}$ $2y=\sqrt{3}\left(4\right)$

$y=2\sqrt{3}$

Centroid=$(-3,\frac{2}{\sqrt{3}})$