Question

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

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

Answer 1

Answer: (C)

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

Given equation of parabola is
$y^2=4x$
The equation of any normal at P($t^2$,2t) is given as
$y=-tx+2t+t^3$
Since it passes through the point (15, 12)
$\Rightarrow 12=-15t+2t+t^3$
$\Rightarrow t^3-13t-12=0$
$\Rightarrow (t+1)(t^2-t-12)=0$
$\Rightarrow t=-1,-3,4$
Therefore, the co-normal points are (1, -2), (9, -6), (16, 8).
Therefore, centroid is $(\frac{26}{3},0)$.