The angle between the tangents drawn from the point (1 ,2 ) to the ellipse 3x^2+2y^2=5 is

Question

Accepted Answer

Correct answer: A. (a) Given Ellipse, 3x^2+2y^2-5=0 Then, equations of pairs at tangents drawn to the êllipse S from the point (1 ,2 ). SS_1=T^2(3x^2+2y^2-5)(3 (1 )^2-2 (2 )^2-5)= (3x (1 )+2y (2 )-5 )^2(3x^2+2y^2-5) (6 )= (3x+4y-5 )^218x^2+12y^2-30=9x^2+16y^2+15+14 xy-40y-30 9x^2-4y^2-24 xy+30x+40y-55=0 Now, the angle between these lines is given by tan ⁡ = 2 h^2- aba+b tan ⁡ = 2 144+365= 2 1805= 12 55 ∴ = tan^-1 ⁡ ( 12 55 )

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

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