Let $a^2 > b^2$ , $s > -b^2$ and $a^2 > t > b^2$ . Show that each of the ellipses

(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: $$\frac{x_0x}{a^2+s}+\frac{y_0y}{b^2+s} = 1, \quad \frac{x_0x}{a^2-t}-\frac{y_0y}{t-b^2} = 1$$ Solving these equations for $y$ shows that the slopes of the tangents are $$m_1 = -\frac{b^2+s}{a^2+s}\!\cdot\!\frac{x_0}{y_0}, \quad m_2 = -\frac{t-b^2}{a^2-t}\!\cdot\!\frac{x_0}{y_0},$$ and thus their product is

(3)

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