Help: Perpendicular project is regular surface?

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SUMMARY

The discussion focuses on demonstrating that the perpendicular projection of the center (0,0,0) of the ellipsoid defined by the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1\) onto its tangent planes forms a regular surface. The tangent plane is derived using the gradient of the function, resulting in the equation \(xx_0/a^2 + yy_0/b^2 + zz_0/c^2 = 1\). The participants emphasize verifying that the projection satisfies the equation \((x^2+y^2+z^2)/2 = a^2x^2 + b^2y^2 + c^2z^2\), with some confusion regarding the division by 2.

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Help: Perpendicular project is regular surface??

(edit found a better way of showing this)

Homework Statement



Hi

I have this problem here which is causing me trouble.

Show that the perpendicular projection of the center (0,0,0) of the ellipsoid
\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1

onto its tangent planes constitutes a regular surface given by

{(x,y,z) \in R^3; (x^2+y^2+z^2)/2 = a^2x^2 + b^2y^2 + c^2z^2}-{(0,0,0)}

What can do here is arrive at tangent plane


The Attempt at a Solution



First I find the tangent

gradF(x0,y0,z0) = <2x0/a^2, 2y0/b^2, 2z0/c^2>

Which gives us a tangentplane

2x0/a^2*(x-x0)+2y0/b^2(y-y0)+2z0/c^2(z-z0) = 0

which by rearrangement gives

xx0/a^2 + yy0/b^2 + zz0/c^2 = 1


the normal is x0/a^2*x = y0/b^2*y = z0/c^2*z


But how do I continue from here?


Best Regards
Alphaboy

Homework Equations

 
Last edited:
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You have a normal n=<x0/a^2, y0/b^2, z0/c^2>.

Can you consider tn and determine what value of t makes tn lie in the plane? Isn't that the projection of the origin into the plane?

Then verify that this point satisfies your desired equation:

(x^2+y^2+z^2)/2 = a^2x^2 + b^2y^2 + c^2z^2

which unfortunately I tried three times, getting two different results, neither of which was exactly this. Are you sure it's divided by 2?
 

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