Question

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

• A. $\left( y + xy _{1}\right)^{2}= p ^{2}\left(1+ y _{1}^{2}\right)$
• B. $\left( y - xy _{1}\right)^{2}= p ^{2}\left(1- y _{1}^{2}\right)$
• C. $\left( y - xy _{1}\right)^{2}= p ^{2}\left(1+ y _{1}^{2}\right)$
• 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}} $
$ \left(\right.$ where $\left.y _{1}=\frac{ dy }{ dx }\right)$
$\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 $
$\Rightarrow \left(y-x y_{1}\right)^{2}=p^{2}\left(1+y_{1}^{2}\right)$