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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    If the tangent at any point \[P\] on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]meets the tangents at the vertices \[A\] and \[A'\] in \[L\] and \[L'\] respectively, then \[AL\cdot \text{ }A'L'\]is equal to

    A) \[a+b\]                

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

    C)  \[{{a}^{2}}\]                                     

    D)  \[{{b}^{2}}\]

    Correct Answer: D

    Solution :

    Let \[P(a\cos \theta ,\,\,b\sin \theta )\] be any point on the ellipse. Then, equation of the tangent at \[P\] is                 \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] It cuts the lines\[x=a\]and\[x=-a\] \[L\left( a,\,\,\frac{b(1-\cos \theta )}{\sin \theta } \right)\]and\[L'\left( -a,\,\,\frac{b(1+\cos \theta )}{\sin \theta } \right)\]respectively. Since, \[A\] and \[A'\] are the vertices of given ellipse. Therefore, coordinates \[A(a,\,\,0)\] and\[B(0,\,\,-a)\]. \[\therefore \]  \[AL=\frac{b(1-\cos \theta )}{\sin \theta }\]and\[AL'=\frac{b(1+\cos \theta )}{\sin \theta }\]


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