Question

The normals to the parabola $y^2=4x$ at the points $P,Q$ and $R$ are concurrent at the point $(15,12)$. The coordinates of the centroid of the triangle $PQR$ are

• A. $(0,\frac{26}{3})$
• B. $\left(\frac{26}{3},0\right)$
• C. $\left(\frac{13}{3},\frac{1}{2}\right)$
• D. $\left(\frac{8}{3},\frac{2}{3}\right)$

Answer 1

Answer: (B)

$\left(\frac{26}{3},0\right)$

The equation of the given parabola is $y^2=4x$
Let $P(t_1^2,2t_1),Q(t_2^2,2t_2),R(t_3^2,2t_3)$ be three points on it.
The equation of the normal to $y^2=4ax$ at the point $(a{t}^2,2at)$ is $y=-tx+2at+a{t}^3$
The normals at $P,Q,R$ are coincident at $(15,12)$
$\Rightarrow t^3-13t-12=0$
The above equation is cubic in $t$ where $t_1,t_2,t_3$ are roots of it.
$\Rightarrow t_1+t_2+t_3=0$
$\Rightarrow t_1^2+t_2^2+t_3^2=(t_1+t_2+t_3{)}^2-2(t_1{t}_2+t_2{t}_3+t_3{t}_1)=26$
$\therefore$ Centroid of the triangle formed by $P,Q,R$ is $\left(\frac{t_1^2+t_2^2+t_3^2}{3},\frac{2t_1+2t_2+2t_3}{3}\right)=\left(\frac{26}{3},0\right)$
Hence, option B is correct