Ellipses and circles are "really" the same: a circle is an ellipse in which the focal distance is $0$. Or, put another way, from among all the ellipses
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
if you are going to choose a "canonical" one to define some functions, the natural choice is to use $a=b=1$, which just leads you back to the unit circle. That is, trying to do "elliptic trigonometric functions" pretty soon either drops you into either an arbitrary choice of parameters, or the regular circular trigonometric functions.
Now, one way in which we can "unify" the circular and the hyperbolic functions is through the complex exponential: the circular trigonometric functions correspond to certain exponentials with purely imaginary arguments, while the hyperbolic ones correspond to purely real arguments: for $t$ a real number,
$$\begin{align*}
\cos t&= \frac{e^{it}+e^{-it}}{2} &\qquad \sin t &= \frac{e^{it}-e^{-it}}{2}\\
\cosh t &= \frac{{e^t}+e^{-t}}{2} & \sinh t &= \frac{e^t-e^{-t}}{2}
\end{align*}$$
Where do these come from? One place to find them is the differential equation
$$y'' + \lambda y = 0.$$
If $\lambda\gt 0$, the solution set is spanned by $\cos(\sqrt{\lambda}\;t)$ and $\sin(\sqrt{\lambda}\;t)$. You get the standard functions by normalizing with $\lambda=1$. If $\lambda\lt 0$, then the solution set is spanned by $\cosh(\sqrt{|\lambda|t})$ and $\sinh(\sqrt{|\lambda|}t)$, and you get the standard functions by normalizing with $\lambda=-1$.
This would suggest looking for any putative "parabolic trigonometric functions" in the only remaining case: $\lambda=0$ (this corresponds to the fact that if you view conic sections as being given by slicing a cone with a plane, you obtain the parabola in the boundary between ellipses and hyperbolas). But the differential equation $y''=0$ has solution space spanned by $1$ and $t$, so that the natural "parabolic functions" are just $1$ and $t$.
The line $\frac{x-2}{2} = \frac{y+4}{3} = \frac{2-z}{5}\,$ can be written as
$$x=2+2t\,,\quad y=-4+3t \,,\quad z=2-5t \,, $$
and the line
$$x = 3 + 4t \,,y = -4 + 6t \,,z = 5 - 10t \,.$$
The two lines have the same direction, since $ v_1=(2,3,-5) $ and $ v_2 = (4, 6,-10)\,, $ where $v_1$ and $v_2$ are the direction vectors of the two lines.
One can get two points lie in the plane. Putting $t=0$ in the equations of the lines gives $p_1=(2,-4,2)$ and $p_2=(3,-4,5)\,,$ which lie in the plane.
Constructing the vector $ v_3=p_2-p_1$ gives $v_3=(1,0,3)\,.$ Now, we can find the normal to the plane by taking the cross product of $v_3$ and $v_1$ or $v_2$.
$$ n = v_3 \times v_2 \,. $$
Once that done, the equation of the plane is given by
$$ n.(X-p_1)=0 \Rightarrow n.( x-2,y-4,z-2 )=0 \,.$$
Best Answer
See the picture.
$F_1F_2=38\Rightarrow c=19.$
The major half-axis is $a=16,\;$ hence $b^2=19^2-16^2=105,$
from where the equation $$\frac{y^2}{256}-\frac{x^2}{105}=1.$$