How to find an ellipse , given 2 passing points and the tangents at them?
Certainly such an ellipse is not uniquely determined. For example, there are lots of ellipses with tangent lines $y=1$ at $(0,1)$ and $y=-1$ at $(0,-1)$: any ellipse $a^2x^2+y^2=1$.
I have not checked that an ellipse with the desired properties always exists. But by a suitable projective transformation we can make the two points nearest neighbours on a pair of parallel lines. After the projection, there is a circle with the desired property, as well as infinitely many non-circular ellipses, and another infinity of hyperbolas. Transform back. It follows that there are infinitely many conics with the desired property.
Here is a complete characterization of all possible ellipses that satisfy the given criteria.
Case 1: The two tangents through the specified points $P$ and $Q$ intersect at a point $O$ distinct from $P$ and $Q$. Then there is an affine transformation mapping $O$, $P$, and $Q$ to $(0,0)$, $(1,0)$, and $(0,1)$ respectively. Under this transformation, any ellipse satisfying the original criteria is mapped to an ellipse passing through $(1,0)$ and $(0,1)$ and tangent to the coordinate axes at those points. All such ellipses are of the form $(x-1)^2+(y-1)^2+2axy = 1$ with $-1 < a < 1$.
Here are some figures from my answer to a duplicate question that I had not realized was a duplicate. The ellipses are shown for $a=\frac12,0,-\frac12$.
Case 2: The two tangents are parallel and neither passes through both points. Then there is an affine transformation mapping $P$ and $Q$ to $(-1,0)$ and $(1,0)$, and the tangents to lines parallel to the $x$-axis. Now there is an infinite family of ellipses given by André Nicolas which satisfy these conditions, namely $ax^2 + y^2 = 1$ with $a > 0$.
Update: See these two previous posts for a way to choose the best ellipse from this one-dimensional family of ellipses, where by "best" we mean the one with the least eccentricity:
- "...how to find the ellipse with minimal eccentricity"
- Roundest ellipse with specified tangents
It's easier than you think.
You can do it to be tangent to any two lines like for example in this picture $AB$ and $BC$:
Take a point like $D$ (in the picture) then reflect it on the lines $AB$ and $BC$ giving u $D$ and $D'_{1}$ (in the pic) then draw the perpendicular bisector of the line $D'D'_{1}$ (which because in triangle $DD'D'_{1}$, $CB$ and $AB$ are perpendicular bisectors and they pass through $B$, the third perpendicular bisector passes through $B$ too).
Now any point on this line can be used, like $E$. Now draw $ED'$ and $ED_{1}'$ til they intersect to the lines $BC$ and $AB$ respectively at points $P$ and $F$ (sorry i didn’t write the $P$ one) then your ellipses focal points are $E$ and $D$ and the two tangent points are $P$ and $F$.
see..... easy!!!!
And you can prove this (in the second pic):
A given line $AB$ and two given points $C$ and $D$. If you draw an ellipse with the two focal points $C$ and $D$ to be tangent to the line $AB$ at, for example, $E$, $E$ would be the point that $CE+DE$ is minimum value.
So how to find this point?
Answer: Reflect the point $D$ on the line and name it $D'$ then the value $CE+DE=CE+D'E$ and its minimum when the three points $C$ and $E$ and $D'$ are collinear (the nearest distance between two points is the straight line).
So why its tangent?
Because if that is not the minimum value for example the point $E'$ then there exists a second point like $E''$ that $CE'+DE'=CE''+DE''$ (because the amount increases continuously if u move the minimum point to either right or left so for every point in the right there exists one on the left) then it is obvious that the ellipse would intersect the line at two points $E'$ and $E''$. Then its not tangent but for the minimum value its only one point.
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