Question

The differential equation of all the straight lines which are at a constant distance of ${}^{\prime }{a}^{\prime }$ from the origin is

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

Answer 1

The correct option is A ${(y-x\left(\frac{dy}{dx}\right))}^2=a^2(1+{\left(\frac{dy}{dx}\right)}^2)$

The equation of lines is,
$\cos \alpha \cdot x+\sin \alpha \cdot y=a$
Differentiating w.r.t. $x$,
$\cos \alpha +\sin \alpha \cdot y^{\prime }=0\Rightarrow \cot \alpha =-y^{\prime }\sin \alpha =\frac{1}{\sqrt{1+(y^{\prime }{)}^2}}\cos \alpha =-\frac{y^{\prime }}{\sqrt{1+(y^{\prime }{)}^2}}\frac{-x{y}^{\prime }}{\sqrt{1+(y^{\prime }{)}^2}}+\frac{y}{\sqrt{1+(y^{\prime }{)}^2}}=a\therefore {(y-x\left(\frac{dy}{dx}\right))}^2=a^2(1+{\left(\frac{dy}{dx}\right)}^2)$