Question

The differential equation for all the straight lines which are at a unit distance from the origin is

• A. ${(y-x\frac{dy}{dx})}^2=1-{\left(\frac{dy}{dx}\right)}^2$
• B. ${(y+x\frac{dy}{dx})}^2=1+{\left(\frac{dy}{dx}\right)}^2$
• C. ${(y-x\frac{dy}{dx})}^2=1+{\left(\frac{dy}{dx}\right)}^2$
• D. ${(y+x\frac{dy}{dx})}^2=1-{\left(\frac{dy}{dx}\right)}^2$

Answer 1

The correct option is C

${(y-x\frac{dy}{dx})}^2=1+{\left(\frac{dy}{dx}\right)}^2$

Since the equation of lines whose distance from origin is unit,
is given by x cos a + y sin a = 1 .........(i)
Differentitate w.r.t.x, we get cos $a+\frac{dy}{dx}sina=\mathrm{0......}\left(ii\right)$
One eliminating the 'a' with the
help of (i) and (ii)
i.e., (i) - x x (ii)
$\Rightarrow sina(y-x\frac{dy}{dx})=1\Rightarrow (y-x\frac{dy}{dx})=coseca......\left(iii\right)Also\left(ii\right)\Rightarrow \frac{dy}{dx}=-cota\Rightarrow {\left(\frac{dy}{dx}\right)}^2=co{t}^{2}a........\left(iv\right)$
Therefore by
$\left(iii\right)and\left(iv\right),1+{\left(\frac{dy}{dx}\right)}^2={(y-x\frac{dy}{dx})}^2.$