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If the straight line $4 y - 3 x + 18 = 0$ cuts the parabola $y^{2} = 64 x$ is P and Q then the angle subtended by PQ at the vertex of the parabola is.

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$\cos^{- 1} (\frac{87}{\sqrt{27409}})$

$\cos^{- 1} (\frac{35}{\sqrt{15210}})$

$\tan^{- 1} (\frac{17}{\sqrt{22409}})$

$\sin^{- 1} (\frac{28}{\sqrt{17309}})$

answer is A.

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Detailed Solution

$\frac{3 x - 4 y}{18} = 1$

Homogenising ; $y^{2} = 64 x (\frac{3 x - 4 y}{18})$

$9 y^{2} = 32 x (3 x - 4 y)$

$96 x^{2} - 128 x y - 9 y^{2} = 0$

$\cos θ = \frac{a + b}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} \Rightarrow θ = \cos^{- 1} (\frac{87}{\sqrt{27409}})$

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