Given the one-parameter family of quadric surfaces ax^2 +…

Given the one-parameter family of quadric surfaces \[ ax^2 + by^2 - \lambda z^2 = c, \] where $\lambda$ is the parameter and $a,b,c$ are real constants. Find the orthogonal trajectories, the one passing through $x=1$, $z^2=y^a$, and determine when the trajectories are quadric surfaces.

Step 1 · PDE for the orthogonal surfaces

Eliminating $\lambda=(ax^2+by^2-c)/z^2$ from $\nabla F$ (with $F=ax^2+by^2-\lambda z^2-c$) gives a normal proportional to $(axz,\,byz,\,-(ax^2+by^2-c))$. Orthogonality with $(-p,-q,1)$ gives \[ axz\,p + byz\,q = ax^2+by^2-c. \]

Step 2 · First integral

\[ \frac{dx}{axz}=\frac{dy}{byz}=\frac{dz}{ax^2+by^2-c}. \] From the first two ratios: $dx/(ax)=dy/(by)$. Integrating: \[ x^b = K\,y^a. \tag{I} \]

Step 3 · Second integral

Integrating the coupled first and third ratios along a characteristic (using $y$ eliminated via (I)) leads, after simplification, to \[ c_2 = z^2 - x^2 - y^2 + \frac{2c}{a}\ln x. \tag{II} \] This is confirmed directly: with $P=axz$, $Q=byz$, $R=ax^2+by^2-c$, \[ P(c_2)_x+Q(c_2)_y+R(c_2)_z = axz\!\left(-2x+\frac{2c}{ax}\right)-2by^2z+2z(ax^2+by^2-c) = 0. \checkmark \]

Step 4 · General solution

\[ z^2 - x^2 - y^2 + \frac{2c}{a}\ln x = \Phi\!\left(\frac{x^b}{y^a}\right), \] where $\Phi$ is an arbitrary differentiable function.

Step 5 · Surface through $x=1$, $z^2=y^a$

Parametrise by $t=y$: $x=1$, $z^2=t^a$. Then $c_1=1/t^a$ and $c_2=t^a-t^2-1$. Expressing $t^a=1/c_1$ and $t^2=c_1^{-2/a}$ gives $\Phi(s)=1/s-s^{-2/a}-1$. Substituting $s=x^b/y^a$: \[ z^2-x^2-y^2+\frac{2c}{a}\ln x = \frac{y^a}{x^b}-\frac{y^2}{x^{2b/a}}-1. \]

Step 6 · Quadric trajectories

For the orthogonal trajectories to be quadric surfaces, the logarithmic term must vanish ($c=0$), and $\Phi$ must reduce to a constant $D$ so that no non-polynomial dependence on $x^b/y^a$ survives. This gives \[ x^2+y^2-z^2 = -D, \] a family of hyperboloids (one-sheeted for $D>0$, two-sheeted for $D<0$) or the cone $x^2+y^2=z^2$ when $D=0$.

Verification

At $x=1$, $z^2=y^a$: both sides of the boxed particular surface equation reduce to $y^a-y^2-1$. $\checkmark$