11th Class Mathematics Conic Sections Question Bank Critical Thinking

  • question_answer On the ellipse \[4{{x}^{2}}+9{{y}^{2}}=1\], the points at which the tangents are parallel to the line \[8x=9y\] are          [IIT 1999]

    A)            \[\left( \frac{2}{5},\ \frac{1}{5} \right)\]                                    

    B)            \[\left( -\frac{2}{5},\ \frac{1}{5} \right)\]

    C)            \[\left( -\frac{2}{5},\ -\frac{1}{5} \right)\]                                 

    D)            \[\left( \frac{2}{5},\ -\frac{1}{5} \right)\]

    Correct Answer: B

    Solution :

     Ellipse is \[\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1\]Þ \[{{a}^{2}}=\frac{1}{4}\], \[{{b}^{2}}=\frac{1}{9}\]            The equation of its tangent is \[4xx'+9yy'=1\]            \\[m=-\frac{4x'}{9y'}=\frac{8}{9}\]Þ\[x'=-2y'\]            and \[4x{{'}^{2}}+9y{{'}^{2}}=1\]Þ\[4x{{'}^{2}}+9\frac{x{{'}^{2}}}{4}=1\]Þ\[x'=\pm \frac{2}{5}\]            When \[x'=\frac{2}{5},\,\,y'=-\frac{1}{5}\]and when \[x'=-\frac{2}{5},\,\,\,y'=\frac{1}{5}\]                    Hence points are \[\left( \frac{2}{5},\,\,-\frac{1}{5} \right)\,\]and \[\left( -\frac{2}{5},\,\,\frac{1}{5} \right)\].

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