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  • The acute angle between the pair of tangents drawn to the ellipse  

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

Option: 1

\mathrm{\tan ^{-1}\left(\frac{16}{7 \sqrt{5}}\right)}


Option: 2

\mathrm{\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)}


Option: 3

\mathrm{\tan ^{-1}\left(\frac{32}{7 \sqrt{5}}\right)}


Option: 4

\mathrm{\tan ^{-1}\left(\frac{3+8 \sqrt{5}}{35}\right)}


Answers (1)

best_answer

\mathrm{2 x^{2}+3 y^{2}=5} \\

\mathrm{\Rightarrow \frac{x^{2}}{\left(\frac{5}{2}\right)}+\frac{y^{2}}{\left(\frac{5}{3}\right)}=1} \\

\mathrm{\Rightarrow a^{2}=\frac{5}{2}, \quad b^{2}=\frac{5}{3}}

Let equation of tangent be

\mathrm{y=m x \pm \sqrt{a^{2} m^{2}+b^{2}}} \\

\mathrm{y=m x \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}}

It passes through (1,3)

\mathrm{\therefore 3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}} \\

\mathrm{\Rightarrow(3-m)^{2}=\frac{5}{2} m^{2}+\frac{5}{3}}

\mathrm{\Rightarrow 9 m^{2}+36 m-44=0} \\

\mathrm{\therefore m_{1}+m_{2}=-4, \quad m_{1} m_{2}=-\frac{44}{9}} \\

\mathrm{\text { Now }\left(m_{1}-m_{2}\right)^{2}=\left(m_{1}+m_{2}\right)^{2}-4 m_{1} m_{2}} \\

                                   = 16+4\left(\frac{44}{9}\right) \\

                                   =\frac{320}{9} \\

\mathrm{\therefore\left|m_{1}-m_{2}\right| =\frac{8 \sqrt{5}}{3}}

Angle between tagents \mathrm{=\tan ^{-}\left(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right)}

                                   \mathrm{=\tan ^{-1}\left(\frac{\frac{8 \sqrt{5}}{3}}{1-\frac{44}{9}}\right) }\\

                                   \mathrm{=\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)}

Hence correct option is 2

Posted by

Riya

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