From the point (15, 12), three normals are drawn to the parabola `y^2=4x` . Then centroid and triangle formed by three co-normals points is `((16)/3,0)` (b) `(4,0)` (c) `((26)/3,0)` (d) `(6,0)`

Options

A. (16/3.0)
B. (4,0)
C. (26/3.0)
D. (6,0)

Answer

C

Explanation

(3) Let the equation of any normal be `y=-tx+2t+t^(3)`.
Since it passes through the point (15,12) we have
`12=-15t+2t+t^(3)`
`ort^(3)-13t-12=0`
One root is -1. Then,
`(t+1)(t^(2)-t-12)=0`
`ort=-1,-3,4`
Therefore, the co-normal points are (1,-2), (9,-6), and (16,8).
Therefore, the centroid is (26/3,0).