A doubt on a calculus problem In Simmons's Calculus with Analytic Geometric, 2nd, pg. 114, there is the following problem:

$\text{47) Sketch the curve}$ $f(x)=\frac{2}{1+x^2}$ $\text{and find the points on it at which the normal passes through the origin.}$

This is how I approached it: 
let $P$ be a point on $f$ and let it have coordenates $(a,b)$. Then $b=\frac{2}{1+a^2}$. If a line is to pass through points $(a,\frac{2}{1+a^2})$ and $(0,0)$, its slope must be $\frac{(\frac{2}{1+a^2})}{a}$. Also, if a line is to be normal to $f$ it must have slope $-\frac{1}{f'(a)}=\frac{(1+a^2)^2}{4a}$. Combining the two equalities, it follows that $a=\pm1$, and we have found two points that satisfy our constraint: $(1,1)$ and $(-1,1)$. 
However, the book says there is another point that satifies it,$(0,2)$. 
So I have actually two questions:


*

*How could I have found that point?

*Shouldn't my approach yield all values of $a$ such that the normal to $f$ at $a$ also passes through $(0,0)$? What is it missing?


Thanks very much.
 A: The derivative is
$$
f'(x)=-\frac{4x}{(1+x^2)^2}
$$
so the normal at $(a,f(a))$ has equation
$$
y-f(a)=\frac{(1+a^2)^2}{4a}(x-a)
$$
provided $a\ne0$. This passes throught the origin if and only if
$$
-\frac{2}{1+a^2}=-a\frac{(1+a^2)^2}{4a}
$$
that is, $(1+a^2)^3=8$, so $1+a^2=2$, therefore $a=\pm1$.
The tangent at $(0,2)$ is $y=2$, and the normal is $x=0$, which passes through the origin as well. This is not covered by the previous considerations.
A: The error you made stemmed from working with the slopes of the normals. This omitted vertical lines. Using a different form of equation for a line would cover those cases as well.
One form that’s convenient for this problem is $\mathbf n\cdot\mathbf x=\mathbf n\cdot\mathbf p$, where $\mathbf p$ is a known point on the line and $\mathbf n$ is a vector orthogonal to the line. The normal to a curve at a point is orthogonal to the tangent at that point, which gives us $$(1,f'(a))\cdot(x,y)=(1,f'(a))\cdot(a,f(a))$$ for the equation of the normal to the curve $y=f(x)$ at $x=a$. For a line through the origin, the right-hand side of this equation is zero, so the normal at $a$ passes through the origin if $$(1,f'(a))\cdot(a,f(a))=a+f(a)f'(a)=0.$$ Substituting our function into this condition produces $$a+{2\over1+a^2}{-4a\over(1+a^2)^2}=a-{8a\over(1+a^2)^3}=0.$$ An obvious solution is $a=0$, and working through the rest of the equation yields $\pm1$, as in the question.
