A) \[ae\]
B) \[be\]
C) \[\frac{ae}{b}\]
D) \[\frac{a{{e}^{2}}}{b}\]
Correct Answer: A
Solution :
[a] Let the circumradius of \[\Delta P{{S}_{1}}{{S}_{2}}\] be R, and area of \[\Delta P{{S}_{1}}{{S}_{2}}\]be \[\Delta \]. \[\therefore \,\,\,\,R=\frac{{{S}_{1}}P\times {{S}_{2}}P\times {{S}_{1}}{{S}_{2}}}{4\Delta }\] \[=\frac{a(1-e\cos \theta )\times a(1+e\cos \theta )\times 2ae}{4\times \frac{1}{2}b\sin \theta \times 2ae}\] \[=\frac{{{a}^{2}}}{2b}\left( {{e}^{2}}\sin \theta +\frac{{{b}^{2}}}{{{a}^{2}}}\cos ec\theta \right)\] \[\therefore \,\,\,\,{{R}_{\min }}=\frac{{{a}^{2}}}{2b}\times \frac{2be}{a}=ae\] \[(\because \,\,\,\,\,{{e}^{2}}\sin \theta +\frac{{{b}^{2}}}{{{a}^{2}}}\cos ec\theta \ge \frac{2be}{a})\]You need to login to perform this action.
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