Mathematical Immaturity

14.21 Miscellaneous Review Exercises

• A circle passes through both foci of an ellipse and is tangent to the ellipse at two points. Find the eccentricity of the ellipse.
• The equation of the ellipse in standard form is \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \end{align*} If the circle passes through the foci of the ellipse, then its radius must have length $ae,$ which means that the equation of the circle is then \begin{align*} \\ \frac{x^2}{a^2e^2} + \frac{y^2}{a^2e^2} &= 1 \end{align*} If the circle is tangent to the ellipse at two points, what does this imply about the points of tangency on the ellipse?
• First, we define our ellipse and circle in terms of the eccentricity $e$ and the distance $d$ from the focus of the ellipse to the directrix. The ellipse can be described by the standard-form equation \begin{align*} \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1 \end{align*} where $a = ed/(1 - e^2)$ and $b^2 = a^2(1 - e^2).$ From Theorem 13.19, we know that its foci are located at $(-ae, 0)$ and $(ae, 0),$ which means that the circle passing through the foci has the standard form equation \begin{align*} \\ \frac{x^2}{a^2e^2} + \frac{y^2}{a^2e^2} &= 1 \end{align*} Now, to find the points of tangency, we must first find the points at which the slope of the circle and the slope of the ellipse are equal. To do so, we take the derivative with respect to $x$ of the ellipse and circle respectively: \begin{align*} \\ \frac{2x}{a^2} + \frac{2y}{b^2}\frac{dy}{dx} &= 0, \quad 2x + 2y\frac{dy}{dx} = 0 \end{align*} And since the circle passes through the foci of the ellipse located at $(-ae, 0)$ and $(ae, 0),$ we know that the circle cannot be tangent to the ellipse when $y = 0.$ As such, we can express the respective slopes of the ellipse and circle as follows: \begin{align*} \\ \frac{dy}{dx} &= -\frac{b^2x}{a^2y}, \quad \frac{dy}{dx} = -\frac{x}{y} \end{align*} Then, setting the two equal, we find that the slope of the circle is parallel to that of the ellipse when $b^2x/a^2 = x,$ or when $e^2x = 0.$ And since we know that $e > 0,$ we can see that the slopes are parallel when $x = 0.$ To show tangency, we must also find the eccentricity $e$ such that the two curves intersect when $x = 0.$ In other words, we wish to find $e$ that satisfies $b^2 = a^2e^2,$ or $e^2 = 1 - e^2.$ This is satisfied by \begin{align*} e &= \frac{1}{2}\sqrt{2} \quad \blacksquare \end{align*}