hyperbolas orthogonal to ellipses∗
pahio†
2013-03-22 1:12:03
Let a2> b2,s > −b2and a2> t > b2. Show that each of the ellipses
x2
a2+s+y2
b2+s= 1 (1)
is an orthogonal curve of every hyperbola
x2
a2−t−y2
t−b2= 1.(2)
Let (x0, y0) be an intersection point of an ellipse (1) and a hyperbola (2).
By polarizing both equations in the point (x0, y0) we get the equations of the
tangents of the curves in this point:
x0x
a2+s+y0y
b2+s= 1,x0x
a2−t−y0y
t−b2= 1
Solving these equations for yshows that the slopes of the tangents are
m1=−b2+s
a2+s·x0
y0
, m2=−t−b2
a2−t·x0
y0
,
and thus their product is
m1m2=−(b2+s)(t−b2)
(a2+s)(a2−t)·x2
0
y2
0
.(3)
On the other hand, the point (x0, y0) satisfies the equation gotten from (1)
and (2) via subtraction:
0 = 1
a2+s−1
a2−tx2
0+1
b2+s−1
t−b2y2
0= (s+t)−x2
0
(a2+s)(a2−t)+y2
0
(b2+s)(t−b2)
∗hHyperbolasOrthogonalToEllipsesicreated: h2013-03-2iby: hpahioiversion: h40704i
Privacy setting: h1i hExamplei h51N20i
†This text is available under the Creative Commons Attribution/Share-Alike License 3.0.
You can reuse this document or portions thereof only if you do so under terms that are
compatible with the CC-BY-SA license.
1
