I am asked to prove the radius of the curvature of the curve $$y^2(a-x)=x^2(a+x)$$ at origin is $4\sqrt{2}a$.
I tried it like this, but $\textbf{I can't get 2 in the answer}$. I check it many times and I broke my head since I can't find where it went wrong. I won't even know whether there is a $\textbf{typo in my book}$ . Please help me to find where it went wrong and correct answer for this question.
On
For the curve $c(x,y)=0$, the curvature is given by $$ k = \frac{|2c_xc_yc_{xy} - c_x^2 c_{yy} - c_y^2 c_{xx}|}{(c_x^2 + c_y^2)^{3/2}} $$ In your example, let $$ c(x,y) = y^2(a-x) - x^2(a+x) $$ Then \begin{align} c_x &= -y^2 - 2ax - 3x^2 \\ c_y &= 2(a-x)y \\ c_{xx} &= -2a -6x \\ c_{xy} &= -2y \\ c_{yy} &= 2(a-x) \end{align} At $(x,y)=(0,0)$, we get \begin{align} c_x &= 0 \\ c_y &= 0 \\ c_{xx} &= -2a \\ c_{xy} &= 0 \\ c_{yy} &= 2a \end{align} So, we get $0/0$. This suggests that there is something strange going on. So, to understand better, let's graph the curve. We get the following for $a=1$:
There are two branches of the curve passing through the origin. Which one should we use to compute curvature? In this particular example, both branches happen to have the same curvature at the origin, but this won't always be the case. So, beware when someone asks you to calculate "the curvature" at a point of a curve given by an implicit equation.
Let's use the branch that slopes upwards from left to right. It has equation $$ y = x \sqrt{\frac{a+x}{a-x}} $$ At $x=0$, we get $$ y' = 1 $$ $$ y'' = \frac{2}{a} $$ and so $$ \rho = a\sqrt2 $$ If your book has some other answer, then your book is wrong.
The RHS of your equation (2) after implicit differentiation should have $+y^2$ instead of $-y^2.$ Also, $y'$ (even in your current formula) is defined at $x=0.$ After all the simplification, you should get $a\sqrt{2}$ as the radius of curvature.