Question

Differential equation of all straight lines which are at a constant distance from the origin is

• A. ${\left(y+x{y}_1\right)}^2=p^2\left(1+y_1^2\right)$
• B. ${\left(y-x{y}_1\right)}^2=p^2\left(1-y_1^2\right)$
• C. ${\left(y-x{y}_1\right)}^2=p^2\left(1+y_1^2\right)$
• D. None of these

Answer 1

Answer: (C)

Any straight lines which is at a constant distance $p$ from the origin is

$x\cos \alpha +y\sin \alpha =p\dots$(i)
Diff. both sides w.r.t.'x', we get
$\cos \alpha +\sin \alpha \frac{dy}{dx}=0$
$\Rightarrow \tan \alpha =-\frac{1}{y_1}$
$($ where $y_1=\frac{dy}{dx})$
$\therefore \sin \alpha =\frac{1}{\sqrt{1+y_1^2}}$;
$\cos \alpha =-\frac{y_1}{\sqrt{1+y_1^2}}$
Putting the value of $\sin \alpha$ and $\cos \alpha$ in (i), we get x.
$\frac{-y_1}{\sqrt{1+y_1^2}}+y\frac{1}{\sqrt{1+y_1^2}}=p$
$\Rightarrow {\left(y-x{y}_1\right)}^2=p^2\left(1+y_1^2\right)$