Q.210. Parabola, Co-normals.

Question 210.
Three normals are drawn to the parabola y^2 = 4x from the point (15, 12).
Find the centroid of the triangle formed by three co-normal points.

Answer 210.
y^2 = 4x
=> 2y dy/dx = 4
=> dy/dx = 2/y
=> Slope of tangent at the parametric point, P = (t^2, 2t) is
m = 1/t
=> slope of normal is m' = - t

=> equation of normal at P is
y - 2t = - t (x - t^2)
=> tx + y = t^3 + 2t

As the normal passes through (15, 12),
15t + 12 = t^3 + 2t
=> t^3 - 13t - 12 = 0
=> t^3 + t^2 - t^2 - t - 12t - 12 = 0
=> t^2 (t + 1) - t (t + 1) - 12 (t + 1) = 0
=> (t + 1) (t^2 - t - 12) = 0
=> (t + 1) (t + 3) (t - 4) = 0
=> t = -1, - 3, 4

These three values of t give three points of intersection of the normal through (15, 12) with the parabola which are the co-normal points. Plugging these values of t in P (t^2, 2t), the three co-normal points are
(1, -2), (9, -6) and (16, 8).

The centroid of the triangle formed by these points is
[(1+9+16)/3, (-2-6+8)/3]
= (26/3, 0).

[Note that for co-normal points, sum of the slopes = 0 => y-coordinate of centroid will always be on the axis of the parabola which is 0 for the given parabola.]

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