Ellipses and hyperbolas are usually defined using two foci, but they can also be defined using a focus and a directrix.
Definitions of conic sections: Let $L$ be the distance to the focus
and let $D$ be the distance to the directrix. Pick a constant $e>0$.
The set of all points with $L=eD$ is
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For instance, suppose that we have a conic section with focus at the origin, directrix at $y=-1$, and eccentricity $e \ne 1$. Then \begin{eqnarray*} \sqrt{x^2 + y^2} &=& e|y+1| \cr x^2 + y^2 & = & e^2 (y^2 + 2 y + 1) \cr x^2 + (1-e^2)y^2 -2e^2 y& = & e^2 . \end{eqnarray*} Notice that the coefficient of $y^2$ in the last equation is positive if $e<1$, giving us an ellipse, and is negative when $e>1$, giving us a hyperbola. After completing the square and applying some more algebraic manipulations, we can put the equation in standard form: $$\left ( \frac{1-e^2}{e^2}\right) x^2 + \left( \frac{(1-e^2)^2}{e^2}\right )\left ( y - \frac{e^2}{1-e^2}\right )^2 = 1.$$
If $e<1$, this is an vertically aligned ("tall and skinny") ellipse with center at $(0,\frac{e^2}{1-e^2})$, with $a = \frac{e}{1-e^2}$, $b=\frac{e}{\sqrt{1-e^2}}$ and $c=\frac{e^2}{1-e^2}$ and eccentricity $e=c/a$. As $e \to 1$, the center and the size of the ellipse both go to infinity.
If $e>1$ the situation is analogous. Our curve is then a hyperbola of with center at $(0,-\frac{e^2}{e^2-1})$, with $a = \frac{e}{e^2-1}$, $b=\frac{e}{\sqrt{e^2-1}}$ and $c=\frac{e^2}{e^2-1}$ and eccentricity $e=c/a$. As $e \to 1$, the center and the size of the hyperbola both go to infinity.
The following video describes — without equations — what happens to a conic section as the eccentricity is increased from 0 to 1, and then beyond 1.