Q118 / 160 · TS EAMCET 2022 (19 Jul Shift 1) · Mathematics

The equation of any member of the family of all the ellipses whose axes are along the coordinate axes satisfies the differential equation

  • Correct answer
    A
    $x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0$
  • B
    $x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y=y^{\prime}$
  • C
    $y^{\prime \prime}+\frac{\left(y^{\prime}\right)^2}{y}-\frac{y}{x}=0$
  • D
    $y^{\prime \prime}+\left(y^{\prime}\right)^2+x^2 y^2=0$

Solution

General equation of ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ Differentiate w.r.t ' $x$ '. $$ \begin{aligned} & \frac{2 x}{a^2}+\frac{2 y}{b^2} \cdot \frac{d y}{d x}=0 \\ & \frac{2 y}{b^2} \cdot \frac{d y}{d x}=-\frac{2 x}{a^2} \\ & \frac{d y}{d x}=\frac{-x}{y} \frac{b^2}{a^2} \\ & \frac{y^{\prime} y}{x}=-\frac{b^2}{a^2} \end{aligned} $$ Again differentiate (i) w.r.t ' $x$ '. $$ \begin{aligned} & \frac{1}{b^2} y \cdot \frac{d^2 y}{d x^2}+\frac{1}{b^2}\left(\frac{d y}{d x}\right)^2=\frac{-1}{a^2} \\ & y y^{\prime \prime}+\left(y^{\prime}\right)^2=\frac{-b^2}{a^2} \\ & y y^{\prime \prime}+\left(y^{\prime}\right)^2=\frac{y y^{\prime}}{x} \\ & x y y^{\prime \prime}+x\left(y^{\prime}\right)^2-y y^{\prime}=0 \end{aligned} $$ So, (a) is correct option.

Hence, the correct answer is option: 'A'.

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