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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Differential Equations

NDA 2016 · Mathematics · Differential Equations

Which one of the following differential equations represents the...

Which one of the following differential equations represents the family of straight lines which are at unit distance from the origin?

Options

  1. A(y−xdydx)2=1−(dydx)2\left(y-x \frac{d y}{d x}\right)^2=1-\left(\frac{d y}{d x}\right)^2
  2. B(y+xdydx)2=1+(dydx)2\left(y+x \frac{d y}{d x}\right)^2=1+\left(\frac{d y}{d x}\right)^2
  3. C(y−xdydx)2=1+(dydx)2\left(y-x \frac{d y}{d x}\right)^2=1+\left(\frac{d y}{d x}\right)^2
  4. D(y+xdydx)2=1−(dydx)2\left(y+x \frac{d y}{d x}\right)^2=1-\left(\frac{d y}{d x}\right)^2

Correct answer

C. (y−xdydx)2=1+(dydx)2\left(y-x \frac{d y}{d x}\right)^2=1+\left(\frac{d y}{d x}\right)^2

Step-by-step solution

y=mx+cy=m x+c (Equation of straight line) dydx=mandmx−y+c=0\frac{d y}{d x}=m and m x-y+c=0 is at unit distance from origin. ∴∣m(0)−(0)+c∣m2+(−1)2=1⇒c=1+m2\therefore \frac{|m(0)-(0)+c|}{\sqrt{m^2+(-1)^2}}=1 \Rightarrow c=\sqrt{1+m^2} Now: [y−xdydx]2=[mx+c−xm]2⇒c2=1+m2\left[y-x \frac{d y}{d x}\right]^2=[m x+c-x m]^2 \Rightarrow c^2=1+m^2 also, [y+xdydx]2=[mx+c+mx]2=[2mx+1+m2]2also,\left[y+x \frac{d y}{d x}\right]^2=[m x+c+m x]^2=\left[2 m x+\sqrt{1+m^2}\right]^2 also,1-\left(\frac{d y}{d x}\right)^2=1−m2and1+(dydx)2=1-m^2 and 1+\left(\frac{d y}{d x}\right)^2=1+m^2\Rightarrow \left[y-x \frac{d y}{d x}\right]^2=1+(dydx)2=1+\left(\frac{d y}{d x}\right)^2
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