Show that the straight line $3x-5y=2z$, $3x+5y=2$ lie entirely on the paraboloid $9x^2-25y^2=4z$

Question

Show that the straight line $3x-5y=2z$, $3x+5y=2$ lie entirely on the paraboloid $9x^2-25y^2=4z$.

Attempt

The equation $$9x^2-25y^2=4z$$ can be written as $$3x-5y=2\lambda z, ~~3x+5y=\frac{2}{\lambda }.$$ If it is a generator, it lies entirely on the given surface.

If a point is given in the question through which the generator passes, then I can easyly find $\lambda $, but it is not given. How to solve it.

Answer 1

Let $(x_0,y_0,z_0)$ lie on the intersection of the planes

\begin{align}3x-5y&=2z\\3x+5y&=2\end{align}

If we multiply the equations we get $$(3x_0-5y_0)(3x_0+5y_0) = 4z_0 \implies 9x_0^2-25y_0^2=4z_0$$

which implies that $(x_0,y_0,z_0)$ lies on $$9x^2-25y^2=4z.$$

Answer 2

The intersection of the given planes is given by

$$\begin{cases}3x-5y-2z&=0\\{}\\3x+5y&=2\end{cases}\implies 10y+2z=2\implies 5y=-z+1\,,\,\,3x=2+z-1=z+1$$

and the line is

$$\left\{\,\left(\frac13(t+1),\,-\frac{t-1}5,\,t\right)\;/\;t\in\Bbb R\,\right\}=\left(\frac13,\,\frac15,\,0\right)+t\left(\frac13,\,-\frac15,\,1\right)\;,\;\;t\in\Bbb R$$

Now, substitute values in the parabolloid's equation:

$$9\left[\frac13(t+1)\right]^2-25\left(\frac{t-1}5\right)^2\stackrel?=4t\iff(t+1)^2-(t-1)^2\stackrel?=4t\iff$$$${}$$

$$4t=4t\;\;\color{green}\checkmark$$

and since the last equality is trivially true we're done.

Answer 3

Use $z$ as independent variable on the line $\ell$. This means solving $$\eqalign{3x-5y&=2z\cr 3x+5y&=2\cr}$$ for $x$ and $y$. The result is $x={1\over3}(1+z)$, $\>y={1\over5}(1-z)$. It follows that in all points of $\ell$ we have $$9x^2-25y^2=(1+z)^2-(1-z)^2=4z\ ,$$ wich proves that every point $(x,y,z)\in\ell$ is lying on the hyperbolic paraboloid given by $9x^2-25y^2=4z$.