Question about Normal to an Ellipsoid?

Omid82

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Hi.

I know (from many books) that normal to an ellipsoid of equation (x^2+y^2)/a^2+(z^2)/b^2=1 at any point will passes the minor axis of that ellipsoid. i just want to know how to prove it?

All Answers are Appreciated /////
 
Hi.

I know (from many books) that normal to an ellipsoid of equation (x^2+y^2)/a^2+(z^2)/b^2=1 at any point will passes the minor axis of that ellipsoid. i just want to know how to prove it?

All Answers are Appreciated /////
This question is beyond what I would consider to be "Geometry and Trig." To find the equation of the normal to a surface, I would find the gradient, which involves partial derivatives.

Surface: x2/a2+y2/a2+z2/b21=0\displaystyle x^2/a^2 + y^2/a^2 + z^2/b^2 - 1 = 0

Normal vector: n=(2x/a2) x^+(2y/a2) y^+(2z/b2) z^\displaystyle \vec{n} = (2x/a^2)\ \hat{x} + (2y/a^2)\ \hat{y} + (2z/b^2)\ \hat{z}

Let (X,Y,Z)\displaystyle (X, Y, Z) be a point on the ellipsoid. The equation of the normal line through that point is

........(2X/a2) (xX)+(2Y/a2) (yY)+(2Z/b2) (zZ)=0\displaystyle (2X/a^2)\ (x - X) + (2Y/a^2)\ (y - Y) + (2Z/b^2)\ (z - Z) = 0

........(X/a2) x+(Y/a2) y+(Z/b2) z[X2/a2+Y2/a2+Z2/b2]=0\displaystyle (X/a^2)\ x + (Y/a^2)\ y + (Z/b^2)\ z - \left[ X^2/a^2 + Y^2/a^2 + Z^2/b^2\right] = 0

........(X/a2) x+(Y/a2) y+(Z/b2) z1=0\displaystyle (X/a^2)\ x + (Y/a^2)\ y + (Z/b^2)\ z - 1 = 0

Can you show that that line intersects the z-axis?
 
This question is beyond what I would consider to be "Geometry and Trig." To find the equation of the normal to a surface, I would find the gradient, which involves partial derivatives.

Surface: x2/a2+y2/a2+z2/b21=0\displaystyle x^2/a^2 + y^2/a^2 + z^2/b^2 - 1 = 0

Normal vector: n=(2x/a2) x^+(2y/a2) y^+(2z/b2) z^\displaystyle \vec{n} = (2x/a^2)\ \hat{x} + (2y/a^2)\ \hat{y} + (2z/b^2)\ \hat{z}

Let (X,Y,Z)\displaystyle (X, Y, Z) be a point on the ellipsoid. The equation of the normal line through that point is

........(2X/a2) (xX)+(2Y/a2) (yY)+(2Z/b2) (zZ)=0\displaystyle (2X/a^2)\ (x - X) + (2Y/a^2)\ (y - Y) + (2Z/b^2)\ (z - Z) = 0

........(X/a2) x+(Y/a2) y+(Z/b2) z[X2/a2+Y2/a2+Z2/b2]=0\displaystyle (X/a^2)\ x + (Y/a^2)\ y + (Z/b^2)\ z - \left[ X^2/a^2 + Y^2/a^2 + Z^2/b^2\right] = 0

........(X/a2) x+(Y/a2) y+(Z/b2) z1=0\displaystyle (X/a^2)\ x + (Y/a^2)\ y + (Z/b^2)\ z - 1 = 0

Can you show that that line intersects the z-axis?

Thank you , but for showing that this line intersects the z axis , i should put x=y=0 ? Thank you again.
 
Thank you , but for showing that this line intersects the z axis , i should put x=y=0 ? Thank you again.
Yes, and then you can see there is a perfectly well-defined solution for z, the point of intersection, as a function of Z, the arbitrary point on the ellipsoid where you constructed the normal. Since the limits of Z are ±b, what are the limits for z? Does that satisfy the theorem that the point of intersection is "on the minor axis" of the ellipsoid?
 
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