From the point (15,12) three normals are drawn to the parabola y2=4 x, then centroid of triangle formed by three co-normal points is -

Question

From the point (15,12) three normals are drawn to the parabola y2=4 x, then centroid of triangle formed by three co-normal points is - (A) ((16/3), 0) (B) (4,0) (C) ((26/3), 0) (D) (6,0)

Tardigrade Admin · Accepted Answer

Let the equation of any normal be y=-t x+2 t+t3 Since it passes through the point (15,12) ∴ 12=-15 t+2 t+t3 i.e. t3-13 t-12=0 ∴ t=-1,-3,4 ∴ The points are (1,-2),(9,-6),(16,8) ∴ Centroid is ((26/3), 0)

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 -

$(\frac{16}{3}, 0)$

(4,0)

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

(6,0)

Let the equation of any normal be $y = - t x + 2 t + t^{3}$
Since it passes through the point $(15, 12)$
$∴ 12 = - 15 t + 2 t + t^{3}$ i.e. $t^{3} - 13 t - 12 = 0$
$∴ t = - 1, - 3, 4$
$∴$ The points are $(1, - 2), (9, - 6), (16, 8)$
$∴$ Centroid is $(\frac{26}{3}, 0)$

Q. From the point (15,12) three normals are drawn to the parabola y2=4x, then centroid of triangle formed by three co-normal points is -

(316​,0)