Ytukyg

I am unsure whether or not my proof for an exercise regarding equidistant points is correct.

Let l be a hyperbolic line $delta gt 0$ and
begin{align*}
E= & ; {p in H^2 mid d(p,l)=delta} \
end{align*}
a) Show that in the hyperboloid model, E is obtained as an intersection of $H^2$ with two affine planes. Find explicit formulas for the two affine planes in form of
begin{align*}
U= & ; {x in R^{2,1} mid langle x,vrangle=r} \.
end{align*}
for some v $in R^{2,1}, v neq 0, r gt 0$.

b)Draw a sketch of l and E in the Poincaré disc model and prove that l and E meet at the boundary of the Poincaré disc.

My proof for a) is this.

Since l is a hyperbolic line, l can be written as
begin{align*}
l= & ; {x in R^{2,1} mid langle x,nrangle=0} \
end{align*}

for a suitable space-like vector n.

We may define the planes:
begin{align*}
U_1= & ; {x in R^{2,1} mid langle x,nrangle=sinh(delta)}~
text{and}\
U_2= & ; {x in R^{2,1} mid langle x,-nrangle=sinh(delta)}
end{align*}

We may also define the curves
begin{align*}
gamma_1= U_1 cap H^2 \
gamma_2= U_2 cap H^2 \
end{align*}

My claim is that $E=gamma_1 cup gamma_2$.

"$subseteq$" Let p $in E$. Since $d(p,l)=delta$.
Then $sinh(d(p,l))=sinh(delta)=|langle p,nrangle|$. Then either
begin{align*}
sinh(d(p,l))&=sinh(delta)=langle p,nrangle ~ text{or}\
sinh(d(p,l))&=sinh(delta)=-langle p,nrangle=langle p,-nrangle.
end{align*}

So either p $in U_1$ or p $in U_2$. This shows E $subseteq y_1 cup y_2$.

"$supseteq$" Let p $in y_1 cup y_2$. Then
begin{align*}
sinh(delta)&=langle p,nrangle ~ text{or}\
sinh(delta)&=langle p,-nrangle
end{align*}

so $sinh(delta)=|langle p,nrangle|=sinh(d(p,l))$ and d(p,l)=$delta$. This shows $gamma_1 cup gamma_2 subseteq$ E.

So the claim is proven.

For b) I did this:
Assume that the claim is false. Then l and E meet inside of the disc. In the hyperboloid model this means we find p $in E cap l$. Using the results from a) we have $langle p,nrangle=0$ and $|langle p,nrangle|=sinh(delta)$. So $0=sinh(delta)$. Since sinh is bijective we get $delta=0$ which is a contradiction.

My questions are:

1) Are my proofs correct or is there something I'm overlooking?

2) I'm not completely sure how to draw the sketch. I know that the line l is represented by a circular arc that intersects the boundary of the disc orthogonally. Since E consists of two curves that have constant distance to l I know that E can be represented by two circular arcs that intersect the boundary in the same two points as l. One arc must be contained between the boundary and l and the other one must lie on the other side of l. But how can we draw this configuration so that the two curves have the same constant distance from l?