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Q.
From the point $(15,12)$ three normals are drawn to the parabola $y^{2}=4 x$, then centroid of triangle formed by three co-normal points is -
Conic Sections
Solution:
Let the equation of any normal be $y=-t x+2 t+t^{3}$
Since it passes through the point $(15,12)$
$\therefore 12=-15 t+2 t+t^{3} $ i.e. $t^{3}-13 t-12=0$
$\therefore t=-1,-3,4$
$\therefore $ The points are $(1,-2),(9,-6),(16,8)$
$\therefore $ Centroid is $\left(\frac{26}{3}, 0\right)$