Question

If for a given curve the distance between the origin and the tangent at an arbitrary point is equal to the distance between the origin and the normal at the same point , then show that the curve equation is $\sqrt{x^2+y^2}=c{e}^{\pm tan-1y/x}$.

Answer 1

Let $P(x,y)$ be any arbitrary point on the curve and slope of tangent at this point be $\frac{dy}{dx}$

Equation of tangent at $P(x,y)$ is

$Y-y=\frac{dy}{dx}(X-x)$

$\Rightarrow X\frac{dy}{dx}-Y=x\frac{dy}{dx}-y$
Therefore distance from origin = $\left|\frac{-x\frac{dy}{dx}+y}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}\right|$

Slope of normal $=-\frac{1}{\frac{dy}{dx}}$
Equation of normal at P is

$Y-y=\frac{-1}{\frac{dy}{dx}}(X-x)$

$\Rightarrow Y\frac{dy}{dx}-y\frac{dy}{dx}=-X+x$
Therefore distance from origin = $\left|\frac{x+y\frac{dy}{dx}}{\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}}\right|$

Now, according to given condition

$|-x\frac{dy}{dx}+y|=|x+y\frac{dy}{dx}|$
either $-x\frac{dy}{dx}+y=x+y\frac{dy}{dx}$ or $x\frac{dy}{dx}-y=x+y\frac{dy}{dx}$

$\Rightarrow (x+y)\frac{dy}{dx}=y-x$ or $(x-y)\frac{dy}{dx}=x+y$

$\Rightarrow \frac{dy}{dx}=\frac{y-x}{y+x}$ or $\frac{dy}{dx}=\frac{x+y}{x-y}$

which are homogeneous differential equations in $x$ and $y$
Put $y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v+x\frac{dv}{dx}=\frac{v-1}{v+1}$ or $v+x\frac{dv}{dx}=\frac{1+v}{1-v}$

$\Rightarrow x\frac{dv}{dx}=\frac{v-1}{v+1}-v$ or $x\frac{dv}{dx}=\frac{1+v}{1-v}-v$

$\Rightarrow x\frac{dv}{dx}=\frac{-v^2-1}{v+1}$ or $x\frac{dv}{dx}=\frac{1+v^2}{1-v}$

Separating variables and integrating

$\Rightarrow \int \frac{v+1}{1+v^2}dv=\int \frac{-dx}{x}$ or $\int \frac{v-1}{1+v^2}dv=\int \frac{-dx}{x}$

$\Rightarrow \int \frac{v}{1+v^2}dv+\int \frac{1}{1+v^2}dv=\int \frac{-dx}{x}$ or $\int \frac{v}{1+v^2}dv-\int \frac{1}{1+v^2}dv=\int \frac{-dx}{x}$

$\Rightarrow \frac{1}{2}\ln |1+v^2|+{\tan }^{-1}v=-\ln x+\ln c$ or $\frac{1}{2}\ln |1+v^2|-{\tan }^{-1}v=\ln c-\ln x$

Hence solution will be
$\frac{1}{2}\ln |1+v^2|+\ln x=\pm {\tan }^{-1}v+\ln c$

$\ln \frac{x\sqrt{1+v^2}}{c}=\pm {\tan }^{-1}v$

$x\sqrt{1+v^2}=c{e}^{\pm ta{n}^{-1}v}$
$\Rightarrow \sqrt{x^2+y^2}=c{e}^{\pm {\tan }^{-1}y/x}$