Find the length of the curve $(x-y)^2=a(x+y)$ , $x^2-y^2=\frac98 z^2$ from the point $O(0,0,0)$ to $A(x_0,y_0,z_0)$

Question

Find the length of the curve $(x-y)^2=a(x+y)$ , $x^2-y^2=\frac98 z^2$ from the point $O(0,0,0)$ to $A(x_0,y_0,z_0)$ $(z_0 \gt0)$

I have the solution in my book but what I'm struggling with is understanding the question.

Can somebody explain it to me what am I exactly finding?

Answer 1

Hint.

The aspect of a given curve does not depend of an isotropic coordinates transform. So making $u = x + y$ and $v = x - y$ we have

$$ v^2- a u = 0\\ u v -b z^2 = 0 $$

now we can find

$$ u = u(z)\\ v = v(z) $$

so the representation can be done with a parametric plotting of $(u(z),v(z),z)$ as follows

sols = Solve[{v^2 == a u, u v == b z^2}, {u, v}]
ParametricPlot3D[({u, v, z} /. sols[[1]] /. {a -> 1, b ->9/8}), {z, 0, 2}, PlotStyle -> {Thick, Blue}]

The curve length can be calculated as

$$ l = \int_{z=0}^{z=z_0}\sqrt{1+u'^2+v'^2}dz $$

NOTE

See below the comment by GEdgar.