Question

Show that the curve such that the distance between the origin and the tangent at an arbitrary point is equalto the distance between the origin and the normal at the same point, $\sqrt{x^2+y^2}=c{e}^{\pm ta{n}^{-1\frac{y}{x}}}$

Answer 1

Equation of curve$\Rightarrow \sqrt{x^2+y^2}=C{e}^{\pm {\tan }^{-1}\left(\frac{y}{x}\right)}$.

Now taking log on both the sides, we get

$\frac{1}{2}\log (x^2+y^2)=\log C\pm {\tan }^{-1}\left(\frac{y}{x}\right)$

Differentiating on both sides, we get

$\frac{2x+2y\frac{dy}{dx}}{2(x^2+y^2)}$$=\pm \frac{x^2}{x^2+y^2}\times \frac{(x\frac{dy}{dx}-y)}{x^2}$

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

Now using positive sign,

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

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

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

Now using negative sign,

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

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

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

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

and let arbitrary point be$(a,b)$

$\frac{dy}{dx}=\frac{a+b}{a-b}=$Slope of Tangent

Equation of line$\Rightarrow (y-b)=\left(\frac{a+b}{a-b}\right)(x-a)$

$ay-by-ab+b^2=ax+bx-a^2-ab$

$(a-b)y-(a+b)x+b^2+a^2=0$

Comparing with $Ax+Bx+C=0$, we get

$A=a-b,B=-(a+b),C=b^2+a^2$

Distance from origin$=\frac{C}{\sqrt{A^2+B^2}}$$=\frac{b^2+a^2}{\sqrt{2(a^2+b^2)}}$$=\sqrt{\frac{a^2+b^2}{2}}$

Now slope of tangent$=\frac{a+b}{a-b}$

Slope of normal$\Rightarrow (y-b)=\left(\frac{b-a}{b+a}\right)(x-a)$

$by+ay-b^2-ab=bx-ax-ab+a^2$

$(a+b)y+(a-b)x-(a^2+b^2)=0$

Distance from origin=$\mid \frac{-(a^2+b^2)}{\sqrt{(a+b{)}^2+(a-b{)}^2}}\mid$=$\sqrt{\frac{a^2+b^2}{2}}$

We can clearly see that Equation I$=$Equation II.