PlanetMath: hyperbolas orthogonal to ellipses

(more info)

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hyperbolas orthogonal to ellipses

(Example)

Let , and . Show that each of the ellipses

$\displaystyle \frac{x^2}{a^2+s}+\frac{y^2}{b^2+s} = 1$

(1)

is an orthogonal curve of every hyperbola

$\displaystyle \frac{x^2}{a^2-t}-\frac{y^2}{t-b^2} = 1.$

(2)

Let $(x_0,\,y_0)$ be an intersection point of an ellipse (1) and a hyperbola (2). By polarizing both equations in the point $(x_0,\,y_0)$ we get the equations of the tangents of the curves in this point:

$\displaystyle \frac{x_0x}{a^2+s}+\frac{y_0y}{b^2+s} = 1, \quad \frac{x_0x}{a^2-t}-\frac{y_0y}{t-b^2} = 1$

Solving these equations for

shows that the slopes of the tangents are

$\displaystyle m_1 = -\frac{b^2+s}{a^2+s}\!\cdot\!\frac{x_0}{y_0}, \quad m_2 = -\frac{t-b^2}{a^2-t}\!\cdot\!\frac{x_0}{y_0},$

and thus their product is

$\displaystyle m_1m_2 = -\frac{(b^2+s)(t-b^2)}{(a^2+s)(a^2-t)}\!\cdot\!\frac{x_0^2}{y_0^2}.$

(3)

On the other hand, the point $(x_0,\,y_0)$ satisfies the equation gotten from (1) and (2) via subtraction:

$\displaystyle 0 = \left(\frac{1}{a^2+s}-\frac{1}{a^2-t}\right)x_0^2+\left(\frac... ... = (s+t)\left(\frac{-x_0^2}{(a^2+s)(a^2-t)}+\frac{y_0^2}{(b^2+s)(t-b^2)}\right)$

Since $s\!+\!t$ cannot be 0, the second factor of the abobe product must vanish, which implies the proportion equation

$\displaystyle \frac{x_0^2}{(a^2+s)(a^2-t)} = \frac{y_0^2}{(b^2+s)(t-b^2)}.$

Utilising this in the equation (3) yields the condition of orthogonality

$\displaystyle m_1m_2 = -1,$

for the tangents, which means that the ellipse and the hyperbola intersect orthogonally.

$\begin{pspicture}(-5,-4)(5,4) \psaxes[Dx=9,Dy=9]{->}(0,0)(-4.5,-3.5)(4.5,3.5) \r... ...sdots[linecolor=red](+1.118,0)(-1.118,0) \par \rput(0,-4.5){$ $} \end{pspicture}$

Note. Both the ellipses (1) and the hyperbolas (2) have the common foci $(\pm\sqrt{a^2\!-\!b^2},\,0)$ , being thus confocal.