Show that the feet of the normals from the point ($\alpha,\beta,\gamma$) to the paraboloid $x^2+y^2=2az$ lie on a sphere

Question

Show that the feet of the normals from the point ($\alpha,\beta,\gamma$) to the paraboloid $x^2+y^2=2az$ lie on the sphere $y^2+x^2+z^2-z(\alpha+\gamma)-\frac{y}{2\beta}(\alpha^2+\beta^2) = 0$, when $\beta \ne 0$

Also show that center of the circle is $\left(\dfrac{\alpha}{4},\dfrac{\beta}{4},\dfrac{a+\gamma}{2}\right)$

MY ATTEMPT

The normal at the point ($\alpha$,$\beta$,$\gamma$) to the paraboloid x²+ y² = 2az is the straight line $\frac{x-\alpha}{\alpha}$=$\frac{y-\beta}{\beta}$=$\frac{z-\gamma}{-a}$

If it passes through the point (f, g, h), then we have $\frac{f-\alpha}{\alpha}$=$\frac{g-\beta}{\beta}$=$\frac{h-\gamma}{-a}$=$\lambda$ (say)

Therefore

$\alpha$=$\frac{f}{1+\lambda}$, $\beta$=$\frac{g}{1+\lambda}$, $\gamma$=h+a$\lambda$

But the point ($\alpha$,$\beta$,$\gamma$) lies on the paraboloid; so we have

($\frac{f}{1+\lambda}$)^2 + ($\frac{g}{1+\lambda}$)^2 = 2a (h + λa)

I can't understand how to obtain the equation of circle from this cubic equation.

Answer 1

Using the circular symmetry of the problem, we can reduce its dimension from $3$ to $2$. Thus consider instead of the paraboloid given the parabola whose equation is

$ 2 a y = x^2 $

A foot of the perpendicular from $(x_0, y_0)$ is the point $(x, y)$ that satisfies two equations

1. $2 a y = x^2 $

2. The direction of the normal at $(x,y)$ is along the vector $(x - x_0, y-y_0) $

The normal vector is given by $( 2 x, - 2 a )$, so we must have

$ (2 x , -2 a ) = K (x - x_0, y- y_0 ) $

We can eliminate $K$ from this vector equation by dividing the $y$ component of both sides by their respective $x$ components, and this gives

$ \dfrac{- a}{x} = \dfrac{ y- y_0 }{x - x_0} $

Cross multiplying then gives

$ x (y - y_0) + a(x - x_0) = 0 $

The trick to apply here is to multiply through by $x$

$ x^2 ( y - y_0) + a x (x - x_0 )= 0 $

Using $ 2 a y = x^2 $ , we get

$ 2 a y^2 - 2 a y y_0 + a x^2 - a x x_0 = 0 $

Divide through by $a$

$ x^2 + 2 y^2 - x x_0 - 2 y y_0 = 0 $

Now write $x^2 $ as $ 2 x^2 - x^2 = 2 x^2 - 2 a y $, then

$ 2 x^2 + 2 y^2 - x x_0 - 2 y ( y_0 + a ) = 0$

Divide through by $2$

$ x^2 + y^2 - \dfrac{1}{2} x x_0 - y ( y_0 + a ) = 0 $

Complete the square

$ (x - \dfrac{1}{4} x_0 )^2 + ( y - \dfrac{1}{2} ( y_0 + a) )^2 = \dfrac{x_0^2}{16} + \dfrac{ (y_0 + a)^2 }{4} $

And this is a circle with center $( \dfrac{x_0}{4} , \dfrac{ y_0 + a }{2} ) $

Having obtained this result, and using symmetry, it follows that if we have a paraboloid, and a point $(\alpha, \beta, \gamma )$, then effectively this equivalent to the parabola case, with

$ x_0 = \sqrt{\alpha^2 + \beta^2} , y_0 = \gamma $

Define the angle of rotation of the point $P = (x_0, 0, y_0)$ by an angle $\phi = \text{atan2}(\alpha, \beta)$, i.e. $\cos \phi = \dfrac{\alpha}{\sqrt{\alpha^2 + \beta^2}} $ and $ \sin \phi =\dfrac{\beta}{\sqrt{\alpha^2 + \beta^2}} $

Then rotate the point $P$ by the angle $\phi$ about the $z$ axis. It is straightforward to verify that the rotation of the two-dimensional center $P$ results in the three-dimensional center $(\dfrac{\alpha}{4} , \dfrac{\beta}{4}, \dfrac{\gamma + a} {2} )$