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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 14

  • question_answer
    If the relation between sub-normal SN and sub-tangent ST at any point S on the curve; \[b{{y}^{2}}={{(x+a)}^{3}}\] is \[p(SN)=q{{(ST)}^{2}},\]then the value of \[p/q\]is

    A) \[8a/27\]                      

    B) \[27/8b\]   

    C) \[8b/27\]                      

    D) \[8/27\]

    Correct Answer: C

    Solution :

    Here given curve, \[b{{y}^{2}}={{(x+a)}^{3}}\]
    Differentiating both the sides, we get \[2by\frac{dy}{dx}=3{{(x+a)}^{2}}\Rightarrow \frac{dy}{dx}=\frac{3{{(x+a)}^{2}}}{2by}\]
    \[\therefore \]  length of subnormal is, \[SN=y\frac{dy}{dx}=\frac{3}{2}\frac{{{(x+a)}^{2}}}{b}\] and length of subtangent is, \[ST=y.\frac{dx}{dy}=\frac{2b{{y}^{2}}}{3{{(x+a)}^{2}}}\]
    \[\therefore \,\,p(SN)=q{{(ST)}^{2}}\]
    \[\Rightarrow \,\,\frac{p}{q}=\frac{{{(ST)}^{2}}}{(SN)}=\frac{8}{27}\frac{{{b}^{3}}{{y}^{4}}}{{{(x+a)}^{6}}}=\frac{8b}{27}\,\,\left( \because \,\,\frac{{{b}^{2}}{{y}^{4}}}{{{(x+a)}^{6}}}=1 \right)\]


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