Question

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

• A. $ {{\tan }^{-1}}\left( \frac{12}{5} \right) $
• B. $ {{\tan }^{-1}}(6\sqrt{5}) $
• C. $ {{\tan }^{-1}}\left( \frac{12}{\sqrt{5}} \right) $
• D. $ {{\tan }^{-1}}(12\sqrt{5}) $

Answer 1

Answer: (C)

Equation of tangent at $ ({{x}_{1}},\,\,{{y}_{1}}) $ to the ellipse $ \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1 $ is $ \frac{x\,\,{{x}_{1}}}{{{a}^{2}}}+\frac{y\,\,{{y}_{1}}}{{{b}^{2}}}=1 $ ..(i)
and Equation of pair of tangent to an ellipse from
$ ({{x}_{1}},\,\,{{y}_{1}}) $ is $ \left( \frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}-1 \right)\left( \frac{x_{1}^{2}}{{{a}^{2}}}+\frac{y_{1}^{2}}{{{b}^{2}}}-1 \right) $
$ ={{\left( \frac{x{{x}_{1}}}{{{a}^{2}}}+\frac{y{{y}_{1}}}{{{b}^{2}}}-1 \right)}^{2}} $ .. (ii)
and angle $ \theta $ between them $ ={{\tan }^{-1}}\frac{2\sqrt{{{h}^{2}}-ab}}{a+b} $ ... (iii)
Given equation of ellipse
$ 3{{x}^{2}}+2{{y}^{2}}=5 $ and point $ (1,\,\,2) $ .
It can be rewritten as, $ \frac{{{x}^{2}}}{5/3}+\frac{{{y}^{2}}}{5/2}=1 $
$ \therefore $ Equation of tangent at $ (1,\,\,2) $ is $ \frac{x}{5/3}+\frac{2y}{5/2}=1 $
$ \Rightarrow $ $ 3x+4y=5 $ [from (ii)]
$ \therefore $ Joint equation of tangents $ \left( \frac{{{x}^{2}}}{5/3}+\frac{{{y}^{2}}}{5/2}-1 \right)\left( \frac{1}{5/3}+\frac{4}{5/2}-1 \right) $
$ ={{(3x+4y-5)}^{2}} $ (from (ii))
$ \Rightarrow $ $ 9{{x}^{2}}-4{{y}^{2}}-24xy+30x+40y-30=0 $
$ \therefore $ $ a=9,\,\,b=-4,\,\,h=12,\,\,g=15 $ (By comparing with $ a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0) $
$ \therefore $ $ \theta ={{\tan }^{-1}}\left( \frac{2\sqrt{144+36}}{5} \right)={{\tan }^{-1}}\left( \frac{2\cdot 2\cdot 3\sqrt{5}}{5} \right) $
$ \theta ={{\tan }^{-1}}(12/\sqrt{5}) $