Question

For a parabola $y^2=4x$, if the point of intersection of co-normals is $(15,12)$, then what is the centroid of triangle formed by co-normal points ?

• A. $(4,0)$
• B. $(\frac{26}{3},0)$
• C. $(6,0)$
• D. $(\frac{16}{3},0)$

Answer 1

The correct option is B $(\frac{26}{3},0)$

Given equation of parabola is $y^2=4x\Rightarrow a=1$
$\Rightarrow$ The general form of point on parabola is $P(t^2,2t)$
The equation of normal to the parabola $y^2=4ax$ at point $(a{t}^2,2at)$ is $y=-xt+2at+a{t}^3$
$\Rightarrow$ The equation of normal: $y=-xt+2t+t^3$
Normals pass through the point $(15,12)$
$\Rightarrow 12=-15t+2t+t^3$
$\Rightarrow t^3-13t-12=0$
$\Rightarrow (t+1)(t^2-t-12)=0$
$\Rightarrow (t+1)(t-4)(t+3)=0$
$\Rightarrow t=-1,-3,4$
Substitute the values of $t$ into the general form of point
$\therefore$ Co-normal points are: $(1,-2),(9,-6),(16,8)$
The centroid of the triangle with coordinates
$(a,d),(b,e),(c,f)$ is $(\frac{a+b+c}{3},\frac{d+e+f}{3})$
$\Rightarrow$The centroid of the triangle $(\frac{1+9+16}{3},\frac{-2-6+8}{3})=(\frac{26}{3},0)$