Question

From the point $(15,12),$ three normals are drawn to the parabola $y^2=4x,$ then centroid of triangle formed by three co-normal points is

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

Answer 1

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

We know that the equation of normal for $y^2=4ax$ is $y=-xt+2at+t^3$
Here, $y^2=4x$ parabola is given so $a=1,$
$y=-xt+2t+t^3$
This equation passes through the point $(15,12)$
$\Rightarrow 12=-15t+2t+t^3$
$\Rightarrow t^3-13t-12=0$
$\Rightarrow (t+1)(t^2-t-12)=0$
$t=-1,-3,4$
Therefore the co-normal points can be determined using the general form of point, $P(a{t}^2,2at)$
We get the co-normal points as $(1,-2),(9,-6),(16,8)$
Now we know the formula of a centroid given three points $(a,d),(b,e)$ and $(c,f)$ is $(\frac{a+b+c}{3},\frac{d+e+f}{3})\Rightarrow (\frac{1+9+16}{3},\frac{-2+(-6)+8}{3})$
$\Rightarrow (\frac{26}{3},0).$