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The differential equation representing the family of ellipse having foci eith on the x-axis or on the $$y$$-axis, center at the origin and passing through the point (0, 3) is :

JEE · Math · previous-year question

  1. A.xy y'' + x (y')2 $$-$$ y y' = 0
  2. B.x + y y'' = 0
  3. C.xy y'+ y2 $$-$$ 9 = 0
  4. D.xy y' $$-$$ y2 + 9 = 0correct

Answer

D. xy y' $$-$$ y2 + 9 = 0

Explanation

Equation of ellipse, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$ As ellipse passes through (0, 3) $$\therefore\,\,\,$$ $${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$$ $$ \Rightarrow $$ b2 = 9 $$\therefore\,\,\,$$ Equation of ellipse becomes, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$ Differentiating w.r.t x, we get, $${{2x} \over {a{}^2}}$$ + $${{2y} \over 9}$$ . $${{dy} \over {dx}} = 0$$ $$ \Rightarrow $$ $${x \over {{a^2}}}$$ = $$-$$ $${y \over 9}.{{dy} \over {da}}$$ $$ \Rightarrow $$ $${x \over {{a^2}}} = - {y \over 9}.y'......$$ (1) We got earlier, $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}$$ = 1 $$ \Rightarrow $$ $${x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1$$ putting value of equation (1) here, $$ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1$$ $$ \Rightarrow $$ $$-$$ xyy' + y2 = 9 $$ \Rightarrow $$ xyy' $$-$$ y2 + 9 = 0

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