JEE Main & Advanced Mathematics Conic Sections Question Bank Ellipse

  • question_answer
    If any tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]cuts off intercepts of length h and k on the axes, then \[\frac{{{a}^{2}}}{{{h}^{2}}}+\frac{{{b}^{2}}}{{{k}^{2}}}=\]

    A)            0     

    B)            1

    C)            ?1   

    D)            None of these

    Correct Answer: B

    Solution :

               The tangent at \[(a\cos \theta ,\,b\sin \theta )\]to the ellipse is                    \[\frac{(a\cos \theta )x}{{{a}^{2}}}+\frac{(b\sin \theta )y}{{{b}^{2}}}=1\] or \[\frac{x}{(a/\cos \theta )}+\frac{y}{(b/\sin \theta )}=1\]                    \[\therefore \]Intercepts are,  \[h=\frac{a}{\cos \theta },\,\,k=\frac{b}{\sin \theta }\]Þ\[\frac{{{a}^{2}}}{{{h}^{2}}}+\frac{{{b}^{2}}}{{{k}^{2}}}=1\].          


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