RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    The equation of ellipse whose vertex is (±5,0) and focus is (±4,0), is

    A)  \[9{{x}^{2}}+25{{y}^{2}}=225\]

    B)  \[25{{x}^{2}}+9{{y}^{2}}=225\]

    C)  \[3{{x}^{2}}+4{{y}^{2}}=192\] 

    D)  None of these

    Correct Answer: A

    Solution :

     Given, vertex\[(\pm 5,0)=(\pm \text{ }a,0)\] \[\Rightarrow \] \[a=5\] and focus\[(\pm 4,0)=(\pm \text{ }ae,0)\] \[\Rightarrow \] \[ae=4\] \[\Rightarrow \] \[e=\frac{4}{a}=\frac{4}{5}\] \[\therefore \] \[b=\sqrt{{{a}^{2}}(1-{{e}^{2}})}\] \[\Rightarrow \] \[b=\sqrt{25\frac{(25-16)}{25}}\] \[\Rightarrow \] \[b=\sqrt{9}\] \[\Rightarrow \] \[b=3\] Hence, equation is\[\frac{{{x}^{2}}}{{{(5)}^{2}}}+\frac{{{y}^{2}}}{{{(3)}^{2}}}=1\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{25}+\frac{{{y}^{2}}}{9}=1\] \[\Rightarrow \] \[9{{x}^{2}}+25{{y}^{2}}=225\]


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