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The differential equation satisfied by the system of parabolas y2 = 4a(x + a) is :

JEE · Math · previous-year question

  1. A.$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
  2. B.$$y{\left( {{{dy} \over {dx}}} \right)^2} - 2x\left( {{{dy} \over {dx}}} \right) + y = 0$$
  3. C.$$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$correct
  4. D.$$y\left( {{{dy} \over {dx}}} \right) + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

Answer

C. $$y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

Explanation

$${y^2} = 4ax + 4{a^2}$$ differentiate with respect to x $$ \Rightarrow 2y{{dy} \over {dx}} = 4a$$ $$ \Rightarrow a = \left( {{y \over 2}{{dy} \over {dx}}} \right)$$ So, required differential equation is $${y^2} = \left( {4 \times {y \over 2}{{dy} \over {dx}}} \right)x + 4{\left( {{y \over 2}{{dy} \over {dx}}} \right)^2}$$ $$ \Rightarrow {y^2}{\left( {{{dy} \over {dx}}} \right)^2} + 2xy\left( {{{dy} \over {dx}}} \right) - {y^2} = 0$$ $$ \Rightarrow y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$

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