# Thread: point on an ellipse

1. ## point on an ellipsoid

Thanks

Yes, normal is different to normalized/unit

I've got a somewhat related question about ellipsoids..

$\displaystyle {x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2} - 1 = 0$

If I'm trying to find a point on its surface that is in a direction v what do I need to do?

For a sphere centred at p with radius r it is just:

$\displaystyle {p + r{v \over \| v \|}}$

-=USER WARNED=-.

3. Originally Posted by scorpion007
Thanks

Yes, normal is different to normalized/unit

I've got a somewhat related question about ellipses..

$\displaystyle {x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2} - 1 = 0$

If I'm trying to find a point on its surface that is in a direction v what do I need to do?

For a sphere centred at p with radius r it is just:

$\displaystyle {p + r{v \over \| v \|}}$
As has been said before quite recently for a surface represented by the
equation:

$\displaystyle F(x,y,z)=0$

any normal to the surface is parallel to: $\displaystyle \nabla F$

RonL

4. Perhaps you misunderstood what I was asking or I am misunderstanding your response. I'm not trying to find a normal to the surface but a point on the surface in a specific direction.

For instance the point on a sphere in a given direction can be found with:

$\displaystyle {\vec{p} + r{\vec{v} \over \| \vec{v} \|}}$
where p is the center and v is the direction vector.

In this case it is simple since the sphere has uniform dimensions in all directions.
I need to know how to go about solving this problem for an ellipsoid which can have different dimensions in all axes.

5. Originally Posted by scorpion007

$\displaystyle {p + r{v \over \| v \|}}$
.
I presume $\displaystyle p$ is the vector whose components are the center of the circle?

I presume $\displaystyle v$ is a vector.

I presume $\displaystyle r$ is a scalar.

Correct?
---
Next time represent vectors as,
$\displaystyle \bold{v}\mbox{ or }\vec{v}$

6. Originally Posted by scorpion007
Thanks

Yes, normal is different to normalized/unit

I've got a somewhat related question about ellipsoids..

$\displaystyle {x^2 \over a^2}+{y^2 \over b^2}+{z^2 \over c^2} - 1 = 0$

If I'm trying to find a point on its surface that is in a direction v what do I need to do?

For a sphere centred at p with radius r it is just:

$\displaystyle {p + r{v \over \| v \|}}$
Write $\displaystyle v = (v_x , v_y, v_z).$ Since the center of the ellipsoid is at the origin, you're looking for the intersection of the ellipsoid and the line defined by $\displaystyle (x,y,z) = t(v_x , v_y, v_z).$ So solve $\displaystyle {(tv_x)^2 \over a^2}+{(tv_y)^2 \over b^2}+{(tv_z)^2 \over c^2} - 1 = 0$ for $\displaystyle t,$ which yields

$\displaystyle \hat{t} = \pm \sqrt{ \left({v_x^2 \over a^2}+{v_y^2 \over b^2}+{v_z^2 \over c^2}\right)^{-1} }$

and $\displaystyle \hat{t}v$ is your point on the surface. You can verify this formula gives your same solution for a sphere centered at the origin where $\displaystyle a = b = c = r.$