An ellipse whose axes are parallel to the coordinate axes is tangent to the $x$-axis at $(4, 0)$ and tangent to the $y$-axis at $(0, 1).$ Find the distance between the foci of the ellipse.
I know that the equation of an ellipse whose axes are parallel to the coordinate axes is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$ where $a$ and $b$ are the semi-major and semi-minor axes, respectively. I also know that the distance between the foci of such an ellipse is $2c,$ where $c^2 = a^2 - b^2.$ So, to find the distance between the foci, I need to find $a$ and $b.$ Since the ellipse is tangent to the $x$-axis at $(4, 0),$ I know that $a = 4.$ Since the ellipse is tangent to the $y$-axis at $(0, 1),$ I know that $b = 1.$ Therefore, $c^2 = 4^2 - 1^2 = 15,$ and $2c = 2\sqrt{15}.$ 