Let $K$ be a field of characteristic $0$. Let $a\in K$. Let $E:y^2=x^3+ax$ and $E': y^2=x^3-4ax$. There is an degree 2 isogeny $\phi : E\to E'$ given by $(x,y)\to (\frac{y^2}{x^2},\frac{y(a-x^2)}{x^2})$.
My question is, why is the value $\phi(0,0)$ is $O$(infinite point) ?
This video(https://www.youtube.com/watch?v=USe1SK9pjDY) includes $\phi(0,0)=O$ as definition, by can we derive this by homogenizing the map ? Thank you for your help.
On
As indicated and confirmed by Viktor Vaughn's post, the kernel of this isogeny consists of the points at infinity, $0_E$, together with the point $P=(0,0)\in E$, a cyclic group of order two.
I looked at this same isogeny while working on a recently deleted question, and I want to add the following, as it explains how this isogeny can be constructed starting from the kernel $\langle P\rangle$.
We need to assume that $a\neq0$ for otherwise $E$ isn't an elliptic curve.
If $Q=(x,y)\in E$ is a generic point, we easily see using the point addition procedure on $E$ that $$ P+Q=(\frac a x, -\frac{ay}{x^2}).\tag{1} $$
Let's look at the function field $K(E)=\Bbb{Q}(x,y)$ of $E$. Addition by $P$ naturally gives rise to an automorphism $\sigma$ of $K(E)$ by the recipe $f(Q)\mapsto f(Q+P)$. Clearly $G=\langle\sigma\rangle\simeq C_2$, and it is a natural question to try and identify the subfield of fixed points of $G$. By $(1)$ we have $\sigma(x)=a/x$ and $\sigma(y)=-ay/x^2$, so $$ u=x+\frac ax,\qquad \text{and}\qquad v=y-\frac{ay}{x^2} $$ are elements of the subfield $K(E)^G$ of fixed points. Furthermore, $$ u=x+\frac ax=\frac{x^2+a}x=\frac y{x^2} $$ and $$ v=y-\frac{ay}{x^2}=\frac{y(x^2-a)}{x^2} $$ satisfy the equation (just expand the terms) $$ v^2=u^3-4au\tag{2} $$ of the other elliptic curve $E'$.
The equation $u=x+a/x$ means that $x$ is quadratic over the subfield $K(u,v)$. The equation $u=y/x^2$ then implies that $y\in K(u,v,x)$. As $\Bbb{Q}(u,v)\subseteq K(E)^G$ and (by basic Galois theory) $[K(E):K(E)^G)]=|G|=2$, we have shown that $K(E)^G=\Bbb{Q}(u,v)\simeq K(E')$.
So we have identified $K(E')$ as a subfield of $K(E)$ such that the extension is quadratic. By general algebraic geometry of curves the field inclusion $K(E')\hookrightarrow K(E)$ comes as a dual from a morphism $\Phi:E\to E'$ of curves, and that is your isogeny (followed by point negation on $E'$).
First note that we must have $a \neq 0$ in order for the curves to be nonsingular. Writing the codomain in projective coordinates, we have that $\phi$ is given by \begin{align} \label{proj} \phi(x,y) = [y^2 : y(a-x^2) : x^2] \, . \tag{$*$} \end{align} Let $P = (0,0) \in E$. As suggested in the comments, we can determine $\phi(P)$ by computing the orders of vanishing of $x$ and $y$ at $P$. I claim that $\newcommand{\ord}{\operatorname{ord}} \ord_P(y) = 1$, so $y$ is a uniformizer at $P$, and $\ord_P(x) = 2$. This is apparent geometrically: the line $y = 0$ intersects $E$ transversely at $P$, while the line $x = 0$ is tangent.
We can also see this algebraically. Let $\newcommand{\O}{\mathcal{O}} \O_P$ be the local ring of $E$ at $P$, and let $\newcommand{\m}{\mathfrak{m}} \m = (x,y)$ be its maximal ideal. From the equation for $E$, we have $$ x(x^2 + a) = y^2 \in \m^2 $$ and since $x^2 + a$ is a unit in $\O_P$ (it doesn't vanish at $P$), then $x \in (y^2)$. Thus $\m = (x,y) = (y)$, so $\ord_P(y) = 1$ and $\ord_P(x) = \ord_P(y^2) = 2$.
As you noted, all entries evaluate to $0$ in the expression for $\phi$ in (\ref{proj}). Dividing each entry by the uniformizer $y$, we have $$ [y : (a-x^2) : x^2/y] $$ and since $$ \ord_P(x^2/y) = \ord_P(x^2) - \ord_P(y) = 4 - 1 = 3 \, , $$ then $x^2/y$ evaluated at $P$ is $0$. Thus \begin{align*} \phi(P) = [0 : (a - 0) : 0] = [0 : a : 0] = [0 : 1 : 0] \end{align*} since $a \neq 0$.