JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    If \[3x+4y=12\sqrt{2}\]is a tangent to the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{9}=1\]for some \[a\in R\], then the distance between the foci of the ellipse is                                                             [JEE MAIN Held on 07-01-2020 Evening]

    A) \[2\sqrt{5}\]    

    B) \[~2\sqrt{7}\]

    C) \[4\]                 

    D) \[2\sqrt{2}\]

    Correct Answer: B

    Solution :

    [b] A line \[y=mx+c\] be a tangent to ellipse
    \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\], if \[{{c}^{2}}={{a}^{2}}{{m}^{2}}+{{b}^{2}}\]
    Here, eq. of tangent is: \[4y\,=-\,3x+12\sqrt{2}\]
    \[\therefore \,\,y=-\frac{3}{4}x+3\sqrt{2}\]
    \[\therefore \,\,\,{{\left( 3\sqrt{2} \right)}^{2}}={{a}^{2}}.{{\left( -\frac{3}{4} \right)}^{2}}+9\]
    \[\therefore \,\,{{a}^{2}}=9\times \frac{16}{9}=16\]
    \[\therefore \,\,Eccentricity\,\,of\,\,ellipse\,=e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}\]
    \[\therefore \,\,\,\,Distance\text{ }between\text{ }foci=2ac=2.4.\frac{\sqrt{7}}{4}\]

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