From the point (15, 12) three normals ae drawn to the parabola y2 = 4x

1.	From the point (15, 12) three normals ae drawn to the parabola y2 = 4x, then centroid of triangle formed by three co-normal points is (a) (5, 0) (b) (5, 4) (c) (9, 0) (d) (26/3 , 0)
Answer» Correct option (d) (26/3, 0) Explanation: Let equation of normal be y = –tx + 2t + t³ It passes through (15, 12). So 12 = –15t + 2t + t³ t³ –13t –12 = 0 (t + 1) ( t + 3) (t – 4) = 0 t = –1, –3, 4 Points are (at², 2at) i.e. (1, – 2), (9, – 6), (16, 8) Centroid is (1 + 9 + 16/3 , -2 - 6 + 8/3) = (26/3, 0)

From the point (15, 12) three normals ae drawn to the parabola y2 = 4x, then centroid of triangle formed by three co-normal points is (a) (5, 0) (b) (5, 4) (c) (9, 0) (d) (26/3 , 0)

Answer»

Correct option (d) (26/3, 0)

Explanation:

Let equation of normal be y = –tx + 2t + t³

It passes through (15, 12).

So 12 = –15t + 2t + t³

t³ –13t –12 = 0

(t + 1) ( t + 3) (t – 4) = 0

t = –1, –3, 4

Points are (at², 2at) i.e. (1, – 2), (9, – 6), (16, 8)

Centroid is (1 + 9 + 16/3 , -2 - 6 + 8/3)

= (26/3, 0)

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From the point (15, 12) three normals ae drawn to the parabola y2 = 4x, then centroid of triangle formed by three co-normal points is (a) (5, 0) (b) (5, 4) (c) (9, 0) (d) (26/3 , 0)

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