The intersection between the cylinder $x^2+y^2=1$ and the plane $z=y+3$ forms an ellipse. I want to find its semiaxis.
I have that $$1-x^2-y^2=y+3 \Leftrightarrow x^2+y^2-y=4.$$
completing square of $y^2-y$ I get
$$x^2+\left(y-\frac{1}{2}\right)^2-\frac{1}{4}=4\Leftrightarrow x^2\left(y-\frac{1}{2}\right)^2=\frac{17}{4}.$$
And this doesn't make sense. The answer is $a=1$ and $b=\sqrt{2}.$
The plane makes a $45^\circ$ angle with the height of the cylinder. The semi-major axis is the shortest radius of the ellipse, which is equal to the radius of the cylinder, i.e. $1$. The semi-major axis is $1\div\cos45^\circ=\sqrt{2}$.
Or you may rotate the system by $\pmatrix{x\\y\\z}=\pmatrix{1 & 0 & 0 \\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\\ 0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}}\pmatrix{X\\Y\\Z}$.
The equation of the cylinder becomes $\displaystyle X^2+\left(\frac{Y+Z}{\sqrt{2}}\right)^2=1$.
The equation of the plane becomes $\displaystyle \frac{-Y+Z}{\sqrt{2}}=\frac{Y+Z}{\sqrt{2}}+3$, i.e., $\displaystyle Y=-\frac{3}{\sqrt{2}}$.
Solving, we have the intersection $\displaystyle X^2+\left(\frac{-\frac{3}{\sqrt{2}}+Z}{\sqrt{2}}\right)^2=1$, which is an ellipse with semi-major axis $\sqrt{2}$ and semi-minor axis $1$.
The ellipsis is described by $c(t)=(\cos(t),\sin(t),\sin(t)+3)$. Since the cylinder’s diameter is $2$, one semi axis has length $1$. The cylinder is parallel to the $z$-axis, so we’re looking for points with highest and lowest third coordinate. These are achieved in $t=-\pi/2$ and $t=\pi/2$. The corresponding points are $(0,-1,2)$ and $(0,1,4)$. Since their distance is $2\sqrt2$, the other semiaxis has length $\sqrt2$.
There is no calculation needed to convince you that the the two semiaxes are $a=\sqrt{2}$ and $b=1$. Draw a figure showing the $(y,z)$-plane and the $x$-axis collapsed to the point $O$. The plane $z=y+3$ in space then appears as a $45^\circ$ line $\ell$ which intersects two generating lines $y=\pm1$ of the cylinder, and in space the $x$-axis is orthogonal to $\ell$. The claim then follows from inspection of this figure.
