Let O be the vertex of the parabola x2 = 4y and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2 : 3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1, 2), is :
- 5x - 4y + 3 = 0
- x - 2y + 3 = 0
- 4x - 5y + 6 = 0
- 5x - y - 3 = 0
Answer
(1) 5x - 4y + 3 = 0
Solution
O = (0, 0); take Q = (2t, t2) on x2 = 4y.
P divides OQ in the ratio 2 : 3:
P = ( 2(2t) + 3(0)5, 2t2 + 3(0)5 ) = ( 4t5, 2t25 )
Let P = (h, k): t = 5h4
k = 25 25h216 = 5h28
C: 5x2 = 8y, i.e. S 5x2 - 8y = 0
Chord bisected at (1, 2): T = S1
T 5x(1) - 4(y + 2), S1 = 5(1)2 - 8(2) = -11
5x - 4y - 8 = -11
5x - 4y + 3 = 0
Answer: (1)