As Intelligenti Pauca pointed out it's possible to do it from the rectangular hyperbola. The drawing gets messy for the whole solution, so it will represent only the dilation part:
EDIT: one does not need to construct $f_1$ neither $f_2$.
Draw points $f_1$ and $f_2$: the foci of the rectangular hyperbola whose vertices are the same of $\mathcal H$ aka $V_1$ and $V_2$. To do so, take the midpoint $O$ of $F_1$ and $F_2$ and draw a square with side $OV_1$ and construct a circle centered at $O$ and with radius equals to the diagonal of that square. This circle meets line $F_1F_2$ in $f_1$ and $f_2$.
Let $Q = \ell \cap F_1F_2$ and construct the line $\ell '$ which is the image of $\ell$ under the dilation in the $y$ (vertical) axis of ratio $\frac ab$ (parameters of $\mathcal H$). Let $W$ be one of the meetings of $\ell '$ with the rectangular hyperbola of foci $f_1$ and $f_2$ and vertices $V_1$ and $V_2$, then our solution is simply point $Z$: the meeting of the perpendicular to $F_1F_2$ through $W$ with $\ell$.
To solve the rectangular case all we need is to know about this transformation of circles in rectangular hyperbolas:
Given line $s$ and rectangular hyperbola $\mathcal R$ of vertices $V_1$ and $V_2$ we can construct the meetings of $s$ and $\mathcal R$ by drawing the circle $c$, of diameter $V_1V_2$, and the line $r$, tangent to $c$ in $V_1$. Take a random point $P$ in $s$, let $T = PV_2 \cap r$ and let $P'$ be such that $(V_2,T;P,P')=-1$. Let $H = r \cap s$ and $t = \overleftrightarrow{HP'}$, then $\{X_1,X_2\} = t \cap c$ and so the points $x_1 = V_2X_1 \cap s$ and $x_2 = V_2X_2 \cap s$ are the meetings of $s$ and $\mathcal R$
