Question

The centroid of the triangle formed by the feet of the normals from the point $(h,k)$ to the parabola $y^2+4ax=0,(a>0)$ lies on

• A. X-axis
• B. Y-axis
• C. $x=h$
• D. $y-k$

Answer 1

The correct option is A X-axis

Coordinates of any point on the parabola $y^2=-4ax$ are $(-a{t}^2,2at)$.

Equation of the normal at $(-a{t}^2,2at)$ is $y-xt=2at+a{t}^3$

If the normal passes through the point $(h,k)$, then

$k-th=2at+a{t}^3\Rightarrow a{t}^3+(2a+h)t-k=0$

which is a cubic equation whose three roots $t_1,t_2,t_3$ are the parameters of the feet of the three normals.

$\therefore$ Sum of the roots $=t_1+t_2+t_3=-\frac{coefficientsof{t}^2}{coeffcicientsof{t}^3}=0$

$\therefore$ Centroid of the triangle formed by the feet of the normals

$=(-\frac{a}{3}({t_1}^2+{t_2}^2+{t_3}^3),\frac{2a}{3}(t_1+t_2+t_3))$

$=(-\frac{a}{3}({t_1}^2+{t_2}^2+{t_3}^3),0)$

which, clearly, lies on the x-axis.