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Question: The angle between the pair of tangents drawn from the point (1, 2) to the ellipse \(3x^{2} + 2y^{2} ...

The angle between the pair of tangents drawn from the point (1, 2) to the ellipse 3x2+2y2=53x^{2} + 2y^{2} = 5is

A

tan⁡−1(12/5)\tan^{- 1}(12/5)

B

tan⁡−1(6/5)\tan^{- 1}(6/\sqrt{5})

C

tan⁡−1(12/5)\tan^{- 1}(12/\sqrt{5})

D

tan⁡−1(6/5)\tan^{- 1}(6/5)

Answer

tan⁡−1(12/5)\tan^{- 1}(12/\sqrt{5})

Explanation

Solution

The combined equation of the pair of tangents drawn from (1,2) to the ellipse 3x2+2y2=53x^{2} + 2y^{2} = 5 is

(3x2+2y2−5)(3+8−5)=(3x+4y−5)2(3x^{2} + 2y^{2} - 5)(3 + 8 - 5) = (3x + 4y - 5)^{2} [using SS1=T2\mathbf{S}\mathbf{S}_{\mathbf{1}}\mathbf{=}\mathbf{T}^{\mathbf{2}}]

⇒ 9x2−24xy−4y2+.......=09x^{2} - 24xy - 4y^{2} + ....... = 0The angle between the lines given by this equation is tan⁡θ=2h2−aba+b\tan\theta = \frac{2\sqrt{h^{2} - ab}}{a + b}

Where a=9a = 9, h=−12,b=−4h = - 12,b = - 4 ⇒ tan⁡θ=12/5\tan\theta = 12/\sqrt{5}

⇒θ=tan⁡−1(12/5)\theta = \tan^{- 1}(12/\sqrt{5})