A) \[{{\tan }^{-1}}\left( \frac{12}{5} \right)\]
B) \[{{\tan }^{-1}}(6\sqrt{5})\]
C) \[{{\tan }^{-1}}\left( \frac{12}{\sqrt{5}} \right)\]
D) \[{{\tan }^{-1}}(12\sqrt{5})\]
Correct Answer: C
Solution :
Key Idea: 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 \[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0)\] \[\therefore \]\[\theta ={{\tan }^{-1}}\left( \frac{2\sqrt{144+36}}{5} \right)={{\tan }^{-1}}\left( \frac{2\cdot 2\cdot 3\sqrt{5}}{5} \right)\] \[\theta ={{\tan }^{-1}}(12/\sqrt{5})\]
You need to login to perform this action.
You will be redirected in
3 sec
