Q237 / 270 · NDA 2017 (Phase 1) · Mathematics

What is the differential equation corresponding to $y^2-2 a y+x^2=a^2$ by eliminating $a$ ?

  • Correct answer
    A
    $\left(x^2-2 y^2\right) p^2-4 p x y-x^2=0$
  • B
    $\left(x^2-2 y^2\right) p^2-4 p x y-x^2=0$
  • C
    $\left(x^2+2 y^2\right) p^2-4 p x y-x^2=0$
  • D
    $\left(x^2+2 y^2\right) p^2-4 p x y+x^2=0$ where $\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}$

Solution

$\begin{aligned} & y^2-2 a y+x^2=a^2 \\ & \Rightarrow y^2-2(y)(a)+a^2+x^2=a^2+a^2 \\ & \Rightarrow(y-a)^2+x^2=2 a^2...(i) \end{aligned}$ Dift w.r.t $x$ $\begin{aligned} & 2(y-a) \cdot y^1+2 x=0 \\ & \Rightarrow(y-a) \cdot p=-x \\ & \Rightarrow y-a=\frac{-x}{p}...(ii) \\ & \Rightarrow a=y+\frac{x}{p}...(iii) \end{aligned}$ $\text { (i) } \Rightarrow\left(\frac{-x}{p}\right)^2+x^2=2\left(y+\frac{x}{p}\right)^2$ $\begin{aligned} & =2\left(\frac{(\mathrm{py}+\mathrm{x})^2}{\mathrm{p} 2}\right) \\ & \Rightarrow \frac{\mathrm{x}^2+\mathrm{p}^2 \mathrm{x}^2}{\mathrm{p}^2}=\frac{2\left(\mathrm{p}^2 \mathrm{y}^2+\mathrm{x}^2+2 \mathrm{pxy}\right)}{\mathrm{p}^2} \\ & \Rightarrow \mathrm{x}^2+\mathrm{p}^2 \mathrm{x}^2-2 \mathrm{p}^2 \mathrm{y}^2-2 \mathrm{x}^2-4 \mathrm{pxy}=0 \\ & \Rightarrow\left(\mathrm{x}^2-2 \mathrm{y}^2\right) \mathrm{p}^2-4 \mathrm{pxy}-\mathrm{x}^2=0 \end{aligned}$

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

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