# Coordinates of tangent points from a point outside a spheroid (ellispoid of revolution)

The blue conics in the figure is an ellipse (it would be in a case a circle), that represents the tangency points, from the line drawn from the point $P$ to the ellipsoid surface in all direction, that means, it is the horizon of the point $P$.

My question: How to calculate the coordinates in $x,y,z$ of a point $M$, that lies on the horizon, and it has a bearing (azimuth) $\theta$ from (true) North clockwise?

My input data are the coordinates of P in terms of cartesian coordinates $X_p,Y_p,Z_p$

Thanks for help and tips in advance

Please see figure of geogebra: $x = (N+hgt)*cos(lat)*cos(lon)$ $y = (N+hgt)*cos(lat)*sin(lon)$ $z = ((1-e^2)*N + hgt)*sin(lat)$

With $N = a/sqrt(1 - e^2*sin(lat)^2)$

$hgt:$ height above ellipsoid surface along the normal.

Setting $hgt=0$, that yields the position of $Q$ bcz it has the same coordinates as $P$, just it lies on the surface

• Hint: the plane in which the blue conic lies is the polar of $P$.
– amd
Jul 6 '18 at 23:45
• Do you want this angle measured in the plane of the blue curve or is it a “spherical” angle on the surface of the ellipsoid?
– amd
Jul 7 '18 at 2:09
• Also, where exactly does $O$ lie? Is it the center of the blue conic or is it some other projection of $P$ onto the conic’s plane or the ellipsoid’s surface?
– amd
Jul 7 '18 at 2:23
• @amd this point is $Q$ not $O$, it is on the ellipsoid's srurface, so it has the same geographic coordinates of $P$ with one exception, it has no elevation above ellipsoid surface, this angle $\theta$ is spherical angle $North,Q,M$ on the ellipsoid but excuse me I represented it as if it were in the plane $(North,Q,M)$, so it is really the azimuth due North clockwise, excuse me again Jul 7 '18 at 23:10
• Sorry, but I still don’t understand exactly how $Q$ is related to $P$. Is the line through them normal to the surface, does it pass through the center of the ellipsoid, or something else entirely?
– amd
Jul 10 '18 at 0:39