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Question

$T h e a n g l e b e t w e e n t h e p a i r o f \tan g e n t s d r a w n t o t h e e l l i p s e$ $3 x^{2} + 2 y^{2} = 5 from the point (1, 2) i s$

Moderate

A

$\tan^{- 1} (\frac{12}{5})$
B

$\tan^{- 1} (6 \sqrt{5})$
C

$\tan^{- 1} (\frac{12}{\sqrt{5}})$
D

$\tan^{- 1} (12 \sqrt{5})$

Solution

$\begin{array}{l} {The equation of the pair of tangents is given by SS}_{1} = T^{2} \\ \Rightarrow (3 x^{2} + 2 y^{2} - 5) ({3.1}^{2} + {2.2}^{2} - 5) = {(3 x + 4 y - 5)}^{2} \\ \Rightarrow 9 x^{2} - 4 y^{2} - 24 x y + 40 y + 30 x - 55 = 0 \\ further angle, θ between them can be found by using \\ \tan θ = \frac{2 \sqrt{h^{2} - ab}}{a + b} = \frac{2 \sqrt{{(12)}^{2} - (9) (- 4)}}{9 + (- 4)} = \frac{2 \sqrt{180}}{5} = \frac{12 \sqrt{5}}{5} \\ ∴ θ = \tan^{- 1} \frac{12}{\sqrt{5}} \end{array}$

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