Let L1, L2 be the lines passing through the point P(0, 1) and touching the parabola 9x2 + 12x + 18y - 14 = 0. Let Q and R be the points on the lines L1 and L2 such that the PQR is an isosceles triangle with base QR. If the slopes of the lines QR are m1 and m2, then 16 (m12 + m22 ) is equal to .
Answer
68
Solution
9x2 + 12x + 4 = -18y + 18, i.e. (3x + 2)2 = -18(y - 1)
A line through P, y = mx + 1, gives 9x2 + (12 + 18m)x + 4 = 0; for tangency D = 0: (12 + 18m)2 = 144, so m = 0 or - 43
Thus L1: y = 1 is parallel to the x-axis and L2 is inclined at θ with = - 43.
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Since PQ = PR, the base QR is perpendicular to a bisector of the angle between L1 and L2; the bisectors are inclined at 2 and 90 + 2.
= 2 21 - 2 2 = - 43 gives 2 2 2 - 3 2 - 2 = 0, so 2 = 2 or - 12 (slopes of the two bisectors).
QR is perpendicular to a bisector, so its slope is the negative reciprocal: m1 = - 12, m2 = 2
Therefore 16 (m12 + m22 ) = 16 ( 14 + 4 ) = 68
Answer: 68