2. Sample Papers
  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    Let P be a point on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1(a>b)\] in the 1st or 2nd quadrant whose foci are \[{{S}_{1}}\] and \[{{S}_{2}}\]. Then the least possible value of the circumradius of \[\Delta P{{S}_{1}}{{S}_{2}}\] will be

    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.
You will be redirected in 3 sec spinner