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

Option: 1

\mathrm{\tan ^{-1}\left(\frac{12}{5}\right)} 


Option: 2

\mathrm{ \tan ^{-1}\left(\frac{6}{\sqrt{5}}\right)}


Option: 3

\mathrm{\tan ^{-1}\left(\frac{12}{\sqrt{5}}\right)}


Option: 4

\mathrm{\tan ^{-1}\left(\frac{5}{12}\right)}


Answers (1)

best_answer


Equation of the pair of tangents to the ellipse from the (1,2) is \mathrm{S S_1=T^2}
or  \mathrm{\left(3 x^2+2 y^2-5\right)\left(3.1^2+2 \cdot 2^2-5\right)=(3 x \cdot 1+2 y \cdot 2-5)^2 }or \mathrm{ 9 x^2-4 y^2-24 x y+30 x+40 y-55=0}

\therefore \quad   Angle between the tangents

\mathrm{\theta=\tan ^{-1}\left|\frac{2 \sqrt{\left(h^2-a b\right)}}{a+b}\right|=\tan ^{-1}\left|\frac{2 \sqrt{(144+36)}}{9-4}\right|=\tan ^{-1}\left(\frac{12}{\sqrt{5}}\right) }
Hence (c) is the correct answer.

Posted by

Deependra Verma

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