I got the plot of an ellipsis and a hyperbola using the following code:

plot([sqrt((1^2 - 0.5)*(1 - x^2/(2^2 - 0.5))), -sqrt((1^2 - 0.5)*(1 - x^2/(2^2 - 0.5))), sqrt((1^2 - 2.5)*(1 - x^2/(2^2 - 2.5))), -sqrt((1^2 - 2.5)*(1 - x^2/(2^2 - 2.5)))], x = -2 .. 2, scaling = constrained, color = [blue, blue, red, red])

How can I get the two vectors n1,n2 as shown in the following picture (the intersection point of the curves defines the beginning of the vectors)

There is the following differential equation:

where l(t)=2+t,  g=10 and the goal is to find φ(t). I wrote the following code:

M := diff(phi(t), t);

M1 := diff(M, t);
g := 10;
M2 := diff(l(t), t);
ode := M1 + 2*M2*M/l(t) + g*sin(phi(t))/l(t) = 0;
ics := phi(0) = 1, D(phi)(0) = 1;
l := t -> 2 + t;
ode;
dsovle([ode, ics]);
 

But I don't get any result. How can I solve this differential equation?

Knowing the realations between cartesian ans shperoidal coordinates (x,y,z and v,u,w respectively)

the goal is to take the following graph:

This graphic consists of three planes (One with v=const, let's say v=2, the other one with u=const, let's say u=π/4 and the last one with w=const let's say w=π/4), of three coordinate lines (line v with u=π/4 and w=π/4, line u with v=2 and w=π/4 and finally line w with v=2 and u=π/4). I wrote the following code but an error occurs...

restart;
with(plots);
with(LinearAlgebra);
F := <2*cosh(1)*cos(x), 2*sinh(1)*sin(x)*sin(t), 2*sinh(1)*sin(x)*cos(t)>;
S1 := plot3d(F, x = -Pi/2 .. Pi/2, t = 0 .. 2*Pi, color = "Green");
F, 1.5*Normalize(diff(F, x), 2);
eval(%, {t = Pi/4, x = Pi/4});
A1 := arrow(%, color = red);
V1 := textplot3d(<`+`(`%%`), v^`0`>, align = {'above', 'right'}, font = [Roman, bold, 14]);
F, 1.5*Normalize(diff(F, t), 2);
eval(%, {t = Pi/4, x = Pi/4});
A2 := arrow(%, color = red);
V2 := textplot3d(<`+`(`%%`), u^`0`>, align = {'below'}, font = [Roman, bold, 14]);
G := <2*cosh(x)*cos(Pi/4), 2*sinh(x)*sin(Pi/4)*sin(t), 2*sinh(x)*sin(Pi/4)*cos(t)>;
S2 := plot3d(G, x = 0 .. 100, t = 0 .. 2*Pi, color = "Cyan");
G, 1.5*Normalize(diff(G, x), 2);
eval(%, {t = Pi/4, x = Pi/4});
A3 := arrow(%, color = red);
V3 := textplot3d(<`+`(`%%`), w^`0`>, align = {'above', 'left'}, font = [Roman, bold, 14]);
display(S1, S2, A1, A2, A3, V1, V2, V3, scaling = constrained, axes = framed, labels = [x, y, z]);
 

We have the equation of the following plane in the three-dimension space :

with m1=10, m2=2, C=20 and r1 and r2 given by the following expressions:

with d1=0.6 and d2=0.4. How can we get in Maple the graph of this equation on the x0y plane (z=0) ?

How can I get the follow graph of paraboloidal surfaces (ξ=1,η=2) with the coordinate vectors ξ⁰ , η⁰ , φ⁰ as shown in the picture.

Realation between cartesian coordinates and paraboloidal :

x=1/2 (ξ² - η²), y=ξ η cos(φ), z=ξ η sin(φ)

I wrote the following code:

Student:-VectorCalculus

S1 := PositionVector([1/2*(-eta^2 + 1), eta*cos(phi), eta*sin(phi)], cartesian[x, y, z]):

LH := PlotPositionVector(S1, eta = -2 .. 0, phi = 0 .. 2*Pi, coordcurve = [eta = -2, tangent = true, binormal = true, vectornum = 1], scaling = constrained)

S2 := PositionVector([1/2*(xi^2 - 4), 2*xi*cos(phi), 2*xi*sin(phi)], cartesian[x, y, z]):

LG := PlotPositionVector(S2, xi = -1 .. 1, phi = 0 .. 2*Pi, coordcurve = [xi = 1, tangent = true, vectornum = 1], scaling = constrained)

And I got the two paraboloidal surfaces with the vectors separently.How can I get only one graph with these two surfaces as shown in the picture (with ξ=1, η=2) ?