Let A, B and C be three points on the parabola y2 = 6x and let the line segment AB meet the line L through C parallel to the x-axis at the point D. Let M and N respectively be the feet of the perpendiculars from A and B on L. Then ( AM BNCD )2 is equal to .
Answer
36
Solution
Here 4a = 6 a = 32. Take A(at12, 2at1), B(at22, 2at2), C(at32, 2at3), so L : y = 2at3.
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Chord AB: 2x - (t1 + t2)y + 2at1t2 = 0; putting y = 2at3 gives D with x = a(t1t3 + t2t3 - t1t2)
CD = a|t1t3 + t2t3 - t1t2 - t32| = a|t1 - t3||t2 - t3|
Since L is horizontal, AM = 2a|t1 - t3| and BN = 2a|t2 - t3|
AM BNCD = 4a2|t1 - t3||t2 - t3|a|t1 - t3||t2 - t3| = 4a = 6
Hence ( AM BNCD )2 = 36
Answer: 36