If the slopes of the lines given by the equation $24 x^{3} - a x^{2} y + 26 x y^{2} - 3 y^{3} = 0$ are in G.P., p is the greatest perpendicular distance of the point (1, 1) from these lines, then $a^{2} + 37 p^{2}$ is equal to

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Solution

$24 x^{3} - 3 y^{3} - x y (a x - 26 y) = 0 3 (8 x^{3} - y^{3}) - x y (a x - 26 y) = 0 3 ((2 x)^{3} - y^{3}) - x y (a x - 26 y) = 0$
$a$ must be equal to $52$
$3 [(2 x)^{3} - y^{3}] - x y (52 x - 26 y) = 0 3 [(2 x)^{3} - y^{3}] - 26 x y (2 x - y) = 0 3 (2 x - y) (4 x^{2} + y^{2} + 2 x y) - 26 x y (2 x - y) = 0 (2 x - y) (12 x^{2} + 3 y^{2} + 6 x y - 26 x y) = 0 (2 x - y) (12 x^{2} + 3 y^{2} - 20 x y) = 0 (2 x - y) (6 x - y) (2 x - 3 y) = 0$
Slopes: $\frac{2}{3}, 2, 6$ are in GP with $r = 3$
Greatest distance from $(1, 1) p = ∣ ∣ ∣ \frac{6 - 1}{\sqrt{37}} ∣ ∣ ∣ \Rightarrow p = \frac{5}{\sqrt{37}}$
$a^{2} + 37 p^{2} = {(52)}^{2} + 37 \times \frac{25}{37} a^{2} + 37 p^{2} = 2729$

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If the slopes of the lines given by the equation $24 x^{3} - a x^{2} y + 26 x y^{2} - 3 y^{3} = 0$ are in G.P., p is the greatest perpendicular distance of the point (1, 1) from these lines, then $a^{2} + 37 p^{2}$ is equal to