JCECE Engineering JCECE Engineering Solved Paper-2013

  • question_answer
    If any tangent to the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] intercepts equal lengths 'I' on the axes, then\[l\]is equal to

    A) \[{{a}^{2}}+{{b}^{2}}\]                  

    B) \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]

    C) \[({{a}^{2}}+{{b}^{2}})\]                               

    D)  None of these

    Correct Answer: D

    Solution :

    The equation of tangent to the given ellipse at point \[P(a\cos \theta ,\,\,b\sin \theta )\]is\[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] Intercept of line on the axes are\[\frac{a}{\cos \theta }\]and\[\frac{b}{\sin \theta }\]. Given that,\[\frac{a}{\cos \theta }=\frac{b}{\sin \theta }=l\] \[\Rightarrow \]               \[\cos \theta =\frac{a}{l}\] and        \[\sin \theta =\frac{b}{l}\] \[\Rightarrow \]               \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =\frac{{{a}^{2}}}{{{l}^{2}}}+\frac{{{b}^{2}}}{{{l}^{2}}}=1\] \[\Rightarrow \]               \[{{l}^{2}}={{a}^{2}}+{{b}^{2}}\]    \[\therefore \] \[l=\sqrt{{{a}^{2}}+{{b}^{2}}}\]


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