I'm attempting to draw an ellipse based on two points. For each of these points I have a vector showing the direction the curve of the ellipse should be at this point (I suppose another way of looking at it is that I have 2 tangent lines to the to the ellipse with the intersection point for each).  I know normally 2 points would not be enough to determine the ellipse, but I thought the vectors might make it possible. Ideally I'm trying to calculate the center point and the major and minor axis (I guess either the actual points or the vector from the center).  I'm not quite sure how to proceed with this, or whether it's actually possible, but any help would be greatly appreciated, thanks.  Edit: Added a simple example of what I'm talking about. For the record, the calculation will be in 3D space.  Okay, so the I have the two tangent lines illustrated here, in the form of a point and a (normalized) vector, the point being the intersection point with the ellipse (it's part of the line and the ellipse). By vectors from the center point for the axes I just meant it as an alternative way of finding the minor and major axis points, stupid thing to put in).      Edit 2: Let's assume they're not parallel.
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Please answer to a question , how to find an ellipse which passes the 2 given points and has the given tangents at them. And one related question is that the given condition can decide just one ellipse which satisfies it?  Thank you in advance.  Edit : May be , I can say the first answer (by Mr. André Nicolas) is for the general case. Is there no special case where only a finite set of ellipses satisfies the condition?  Edit : According to the answers and comments , I can compute an arbitrary chosen ellipse for my condition (using the method by Mr.Patrick Da Silva). But there are possibly many others which satisfy my conditions. Am I right?
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How to prove geometrically that if we have a tangent of ellipse with focus F and F' in point P, that tangent is bisector of the angle between a line joining focus F to point P and the line F'P outside the ellipse?
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I have a mesh and in the region around each triangle, I want to compute an estimate of the principal curvature directions. I have never done this sort of thing before and  does not help a lot. Can you describe or point me to a simple algorithm that can help me compute this estimate?  Assume that I know positions and normals of all vertices.
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Suppose you are given 3 circles A,B,C such that A is tangent to B and B is tangent to C. How could you construct a circle (not necessarily all of them) that is tangent to all 3? (If such a circle exists)  I have developed a method of constructing a circle tangent to two tangent circles (pick a point P on their radical axis, draw the tangents from that point to both circles, for each circle construct the line joining the circle center and the tangency point from P, intersect those lines, and draw the circle centered in this intersection and which contains one of the tangency points from P) but I can not figure out how to adapt this idea for 3 circles.
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I am trying a derivation of the polar form of an ellipse using vector notation. Beginning with a definition of an ellipse as the set of points in $\vec{R}^2$ for which the sum of the distances from two points is constant, I have $|\vec{r_1}|+|\vec{r_2}| = c$  Thus, $|\vec{r_1}|^2+|\vec{r_1}||\vec{r_2}|=c|\vec{r_1}|$    Choosing a coordinate system (similar to the one in the diagram), such that $\vec{r_1}=\vec{r}=x\vec{i}+y\vec{j}$, and $\vec{r_2}=\vec{F_2 P}=(x-a)\vec{i}+y\vec{j}$  Noticing that $|\vec{r_1}||\vec{r_2}|=\frac{\vec{r_1}\cdot\vec{r_2}}{\cos(\theta)}=\frac{x^2-ax+y^2}{\cos(\theta)}=\frac{r^2-ax}{\cos(\theta)}$  and using $x=r\cos(\theta)$, we get  $r^2+\frac{r^2}{\cos(\theta)}-ar=cr$, thus I get a polar equation for an ellipse of $r=\frac{a+c}{1+\sec(\theta)}$  whereas I expect $r=\frac{a(1-e^2)}{1+e\cos(\theta)}$. Please tell me where this argument takes the wrong turn!
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Find a vector equation for the line through the point $(a,b,c)$ which is normal to the plane through $P(x_1,y_1,z_1), Q(x_2,y_2,z_2) $and $R(x_3,y_3,z_3)$  What would the process be to answer a question like this?
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Given two vectors $u,v\in\mathbb{R}^3$, find the parametric equations for the plane through the origin parallel to $u$ and $v$.  I'm having a difficult time understanding this question. How can a plane through the origin in $\mathbb{R}^3$ be parallel to two vectors, both of which start from the origin? Surely they all intersect at the origin and thus not parallel?  More concretely, $u=(1,2,3)$ and $v=(-2,1,1)$. The cartesian equation of the plane is $x+7y-5z=0$, which is essentially the solution. But I want to know why.
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can anybody help me to draw this minimal example of an xy axis? I couldn't find anything similar on the forum.  The axis should be black but the grid behind it a shade o gray, dashed. It would be great if the axis were a little bit thicker than the grid also.
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Question: Best way to derive an ellipse formula when given bunch of points (say 60 points which when connected draws an ellipse)?  Background: Using R Momocs library function  which returns confidence ellipse coordinates. Would be nice to abstract those points into a formula.  You can see my ellipse points in the image below;    Any suggestions much appreciated, Thanks
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What is basically a vector ? I am not a starter in learning vectors , have used them in physics and mathematics and done with 1st year of my BS in physics .  But what is a vector basically ? Is it anything that has direction and magnitude ?  or is it something that obeys vector laws , but this statement seems kind of circular ?  How can we go about doing geometry using vectors ? and represent shapes in euclidean geometry using vectors ?
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I have the point $(1,-1)$  and the ellipse $$x^2/9 + y^2/5 = 1 $$  How to find the minimum and maximum distance of the point from the ellipse ?  from exploring the ellipse I know that $$a = 3$$ , $$b =\sqrt{5}$$ $$ c = \sqrt{a^2-b^2} =\sqrt{9-5} = \sqrt{4}=2$$  the eccentricity of the ellipse is $$e=c/a = 2/3 $$ the center is $(0,0)$ and the guides are $$x=3,~~x=-3,~~y=\sqrt{5},~~y=-\sqrt{5}$$  the focus points are : $(2,0)$ and  $(-2,0)$  how from all of that do I find the requested in the question?
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Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ? If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?  Please take a look at this picture:    Actually I know how to determine the arc length of a ellipse . but I want to do the obverse.
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Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$ and pass through $(4,6)$.  I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find the other tangents. Any help is appreciated.
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I'm trying to speed up my workflow by making vector points move in the opposite direction simultaneously when I have two highlighted. For instance, below I highlight the top and bottom of the arrow so they will move at the same time, except I want the bottom point to come up when I'm moving the top point down. Right now, they just move up together or down together. So if point 1 is at Y:20 and point 2 is at Y:10, I want them both to go to Y:15 if I move point 1 down 5 pixels.    What it currently does    What I want it to do
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An entry level confusion about spacetime. I understand that a 4-vector describes a point or event in spacetime. But I've also read (Bertschinger, 1999) that re spacetime "we are discussing tangent vectors that lie in the tangent space of the manifold at each point". If a point/event is described by a single 4-vector, what are all these tangent vectors that lie on the same point? Do they have different co-ordinates to the 'point/event' 4-vector? Could I also ask how a 4-vector 'contains' any sense of direction (I'm thinking here of a vector having direction and magnitude)?
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I have a question regarding the conics section of Spivak calculus and  the way he derives the final equation. I understand everything that he is doing up until:  Now we have to choose coordinate axes in the plane P. We can choose L as the first axis, measuring distances from the intersection Q with the horizontal plane (Figure 5); for the second axis we just choose the line through Q parallel to our original second axis. If the first coordinate of a point in P with respect to these axis is x, then the first coordinate of this point with respect to the original axes can be written in the form $$\alpha x+\beta$$ for some $\alpha$ and $\beta$  Now, I understand everything before that, and why he is changing coordinates from the 'standard' ones, xyz, to coordinates within the plane, so he can express the intersection of the cone and the plane in a equation. But the thing is how does he justify that there is some $\alpha$ and some $\beta$ that satisfy this condition for all x ? He just claims that such values exist, with no proof.  For anyone not familiarized with this chapter, i leave it down here so you can read it.   One of the simplest subsets of this three-dimensional space is the (infinite) cone illustrated in Figure 2; this cone may be produced by rotating a "generaling line," of slope C say, around the third axis.  For any given first two coordinates x and y, the point (x,y,0) in the horizontal plane has distance $$\sqrt{x^2+y^2}$$ from the origin, and thus $$\tag1 (x,y,z)\text{ is in the cone if and only if }z=\pm C\sqrt{x^2+y^2}$$. We can descend from these three-dimensional vistas to the more familiar two-dimnensional one by asking what happens when we intersect this cone with some plane P (Figure 3).  If the plane is parallel to the horizontal plane, there's certainly no mystery--the intersection is just a circle. Otherwise, the plane P intersects the horizontal plane in a straight line. We can make things a lot simpler for ourselves if we rotate everything so that this intersection line points straight out from the plane of the paper, while the first axis is in the usual position that we are familiar with. The plane P is thus viewed "straight on," so that all we see (Figure 4) is its intersection L with the plane of the first and third axes; from this view-point the cone itself simply appears as two straight lines.  In the plane of the first and third axes, the line L can be described as the collection of all points of the form $$(x,Mx+B)$$, where M is the slope of L. For an arbitrary point (x,y,z) it follows that $$\tag2 (x,y,z)\text{ is in the plane }P\text{ if and only if }z=Mx+B.$$ Combining (1) and (2), we see that (x,y,z) is in the intersection of the cone and the plane if and only if $$\tag{$*$} Mx+B=\pm C\sqrt{x^2+y^2}.$$ Now we have to choose coordinate axes in the plane P. We can choose L as the first axis, measuring distances from the intersection Q with the horizontal plane (Figure 5); for the second axis we just choose the line through Q parallel to our original second axis. If the first coordinate of a point in P with respect to these axis is x, then the first coordinate of this point with respect to the original axes can be written in the form $$\alpha x+\beta$$ for some $\alpha$ and $ \beta$. On the other hand, if the second coordinate of the point with respect to these axes is y, then y is also the second coordinate with respect to the original axes.  Consequently, (*) says that the point lies on the intersection of the plane and the cone if and only if $$M(\alpha x+\beta)+B=\pm C\sqrt{(\alpha x+\beta)^2+y^2}.$$ Although this looks fairly complicated, after squaring we can write this as $$\alpha^2C^2y^2+\alpha^2(M^2-A^2)x^2+Ex+F=0$$ for some E and F that we won't bother writing out. Dividing by $\alpha^2$ simplifies this to $$C^2y^2+(C^2-M^2)x^2+Gx+H=0.$$
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From a typical point $P$ inside an ellipse, how many points $Q_i$ on the ellipse have $PQ_i$ normal to the ellipse?  Someone asked me at school many years ago but I don't think I worked it out.