• # question_answer The tangent at a point whose eccentric angle $60{}^\circ$ on the ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\,\,(a>b)$ meet the auxiliary circle at L and M. If LM subtends a right angle at the centre, then eccentricity of the ellipse is- A) $\frac{1}{\sqrt{7}}$                              B) $\frac{2}{\sqrt{7}}$ C) $\frac{3}{\sqrt{7}}$                              D) $\frac{1}{2}$

[B]  Point P will be $(a\cos 60{}^\circ ,b\sin 60{}^\circ )$ Tangent will be $\frac{x\cos 60{}^\circ }{a}+\frac{y\sin 60{}^\circ }{b}=1$ $\frac{x}{2a}+\frac{y\sqrt{3}}{2b}=1$ ?(1) Homogenise ${{x}^{2}}+{{y}^{2}}={{a}^{2}}$ with the help of eqn. (1) We get ${{x}^{2}}+{{y}^{2}}={{a}^{2}}{{\left[ \frac{x}{2a}+\frac{y\sqrt{3}}{2b} \right]}^{2}}$ ${{x}^{2}}+{{y}^{2}}={{a}^{2}}\left[ \frac{{{x}^{2}}}{4{{a}^{2}}}+\frac{3{{y}^{2}}}{4{{b}^{2}}}+\frac{xy\sqrt{3}}{2ab} \right]$ $\frac{3{{x}^{2}}}{4}+{{y}^{2}}\left( 1-\frac{3{{a}^{2}}}{4{{b}^{2}}} \right)-\frac{a\sqrt{3}xy}{2b}=0$ For $90{}^\circ$ at center $\frac{3}{4}+1-\frac{3{{a}^{2}}}{4{{b}^{2}}}=0$ $\frac{3}{4}\,\,\,\frac{{{a}^{2}}}{{{a}^{2}}(1-{{e}^{2}})}=\frac{7}{4}$ $1-{{e}^{2}}=\frac{3}{7}\Rightarrow {{e}^{2}}=\frac{4}{7}\Rightarrow e=\frac{2}{\sqrt{7}}$