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\[=(\pm ae,\,\,0)\], vertices\[\equiv (\pm a,\,\,0)\equiv (0,\,\,0)\] \[\therefore \]\[s=\left| \frac{mae+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|,\,\,\,s'=\left| \frac{-mae+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|\]\[a=\left| \frac{ma+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|,\,\,a'=\left| \frac{-ma+\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|,\]\[c=\left| \frac{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}}{\sqrt{1+{{m}^{2}}}} \right|\] \[\therefore \] \[\frac{ss'-{{c}^{2}}}{aa'-{{c}^{2}}}=\frac{-\frac{{{m}^{2}}{{a}^{2}}{{e}^{2}}}{1+{{m}^{2}}}}{\frac{{{m}^{2}}{{a}^{2}}}{1+{{m}^{2}}}}={{e}^{2}}\]
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