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Let P be any point on the line x - y + 3 = 0 and let A = (3, 4) be a fixed point. A family of lines given by (3 sec(t) + 5 cosec(t)) x + (7 sec(t) - 3 cosec(t)) y + 11(sec(t) - cosec(t)) = 0 passes through a fixed point B for all permissible values of t. If the maximum value of |PA - PB| is expressed as 2 * sqrt(2n) where n is a natural number, find n.

1. 3
2. 4
3. 5
4. 6

Correct answer: 5

Solution

The family of lines can be rewritten as sec(t) * (3x + 7y + 11) + cosec(t) * (5x - 3y - 11) = 0. For this to hold for all t, the concurrent point B must satisfy both 3x + 7y + 11 = 0 and 5x - 3y - 11 = 0 simultaneously. Solving gives B = (1, -2). Then |AB| = sqrt((3-1)² + (4+2)²) = sqrt(40) = 2 * sqrt(10). The line through A and B meets x - y + 3 = 0 at t = 3/2, giving P0 = (4, 7) which lies beyond A. At P0, |P0A - P0B| = |sqrt(10) - 3*sqrt(10)| = 2*sqrt(10) = 2*sqrt(2*5), so n = 5.

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