A) e
B) 1/e
C) \[1/{{e}^{2}}\]
D) \[{{e}^{2}}\]
Correct Answer: D
Solution :
Let the equation of tangent\[y=mx+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}\] Foci\[\equiv (\pm ae,0),vertices\equiv (\pm a,0),C\equiv (0,0)\] \[\therefore \]\[s=\left| \frac{mae+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+m}} \right|,\] \[s'=\left| \frac{-mae+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|\] \[a=\left| \frac{ma+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|\] \[a'=\left| \frac{-ma+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|,c=\left| \frac{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|\] \[\therefore \]\[\frac{ss'-{{c}^{2}}}{aa'-{{c}^{2}}}=\frac{-\frac{{{m}^{2}}{{a}^{2}}{{e}^{2}}}{1+{{m}^{2}}}}{\frac{{{m}^{2}}{{a}^{2}}}{1+{{m}^{2}}}}={{e}^{2}}\]
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