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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Differential Equations

NDA 2017 · Mathematics · Differential Equations

What is the differential equation corresponding to y^2-2 a...

What is the differential equation corresponding to y2−2ay+x2=a2y^2-2 a y+x^2=a^2 by eliminating aa ?

Options

  1. A(x2−2y2)p2−4pxy−x2=0\left(x^2-2 y^2\right) p^2-4 p x y-x^2=0
  2. B(x2−2y2)p2−4pxy−x2=0\left(x^2-2 y^2\right) p^2-4 p x y-x^2=0
  3. C(x2+2y2)p2−4pxy−x2=0\left(x^2+2 y^2\right) p^2-4 p x y-x^2=0
  4. D(x2+2y2)p2−4pxy+x2=0\left(x^2+2 y^2\right) p^2-4 p x y+x^2=0 where p=dydx\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}

Correct answer

A. (x2−2y2)p2−4pxy−x2=0\left(x^2-2 y^2\right) p^2-4 p x y-x^2=0

Step-by-step solution

y2−2ay+x2=a2⇒y2−2(y)(a)+a2+x2=a2+a2⇒(y−a)2+x2=2a2...(i)\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 xx2(y−a)⋅y1+2x=0⇒(y−a)⋅p=−x⇒y−a=−xp...(ii)⇒a=y+xp...(iii)\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} (i) ⇒(−xp)2+x2=2(y+xp)2\text{ (i) } \Rightarrow\left(\frac{-x}{p}\right)^2+x^2=2\left(y+\frac{x}{p}\right)^2=2((py+x)2p2)⇒x2+p2x2p2=2(p2y2+x2+2pxy)p2⇒x2+p2x2−2p2y2−2x2−4pxy=0⇒(x2−2y2)p2−4pxy−x2=0\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}$$
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