Cut in one point on the y-axis on coordinates system is the function $f(x)=4x^2$ with the Coordinate origin $O$. On the function are the points $w, x, y,$ and $z$. The angles $wOx = yOz = 90^o$. Show that $wx$ and $yz$ are cutting each other on the $y$-axis.
Every suggestion is desired. Thanks.
 A: Call $(a,4a^2)$ the coordinates of $w$. Then the segment $Ow$ lies on the line $Y=4aX$. Since $wOx$ is $90^o$, the segment $Ox$ lies on the line $\displaystyle Y=-\frac{1}{4a}X$. So the coordinates of $x$ are $\displaystyle (-\frac{1}{16a}, \frac{1}{64a^2})$, and the segment $wx$ lies on a line whose slope is 
$$\frac{4a^2-\frac{1}{64a^2}}{a+\frac{1}{16a}}=\frac{(256a^4-1)}{4a(16a^2+1)}=\frac{16a^2-1}{4a}$$
Because this line crosses $w$, we get that the intercept $K$ satisfies
$\displaystyle 4a^2=\frac{16a^2-1}{4a}a+K$, which leads to
$$\displaystyle K=4a^2-\frac{16a^2-1}{4}=\frac{1}{4}$$
Thus, the intercept of the segment $wx$ is independent of the initial coordinates chosen for $w$. 
We can therefore apply the same calculations to points $y$ and $z$, calling $(b, 4b^2)$ the coordinates of $y$, and so on. Proceding as above, we get that the segment $yz$ has intercept independent of the initial point coordinates as well, i.e. equal to $\frac{1}{4}$. So the two lines cross each other on the $y$-axis.
