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Two rays aimed towards the point (5, 0), after reflection from outer surface of the hyperbola $9 x^{2} - 16 y^{2} = 144$ at points P and Q intersect at R. If abscissa of points P and Q is 8, then which of the following options is/are correct?

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Angle between asymptotes of the given hyperbola is $\tan^{- 1} \frac{24}{7}$

Area of $Δ P Q R i s 39 \sqrt{3} s q . u n i t s$

Angle between asymptotes of the given hyperbola is $\tan^{- 1} \frac{3}{4}$

Area of $Δ P Q R i s 27 \sqrt{3} s q . u n i t s$

answer is B, D.

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Detailed Solution

$\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \Rightarrow e^{2} = 1 + \frac{9}{16}$

$e = \frac{5}{4} ∴ (\pm a e, 0) = (\pm 5, 0)$

$P (8, 3 \sqrt{3})$

$A = \frac{1}{2} \times 6 \sqrt{3} (5 + 8) = 39 \sqrt{3}$

$θ = 2 \tan^{- 1} (\frac{3}{4}) = \tan^{- 1} (\frac{2. \frac{3}{4}}{1 - \frac{9}{6}}) = \tan^{- 1} (\frac{24}{7})$

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