Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• (a) Prove that a similarity transformation (replacing $x$ by $tx$ and $y$ by $ty$) carries an ellipse with center at the origin into another ellipse with the same eccentricity. In other words, similar ellipses have the same eccentricity.
  (b) Prove also the converse. That is, if two concentric ellipses have the same eccentricity and major axes on the same line, then they are related by similarity transform.
  (c) Prove results corresponding to (a) and (b) for hyperbolas.
• Recall Theorem 13.19: Let $C$ be a conic section with eccentricity $e \neq 1$ and with a focus $F$ at a distance $d$ from a directrix $L.$ If $N$ is a unit normal to $L$ and if $F=eaN,$ where $a = \frac{ed}{1- e^{2}},$ then C is the set of all points $X$ satisfying the equation $$\|X\|^{2} + e^{2}a^{2} = e^{2}(X \cdot N)^{2} + a^{2}$$This equation shows symmetry about the origin since the equation is satisfied by both $X$ and $-X.$ Replacing $X$ by its rectangular coordinates $(x, y)$ and setting $N = \textbf{i},$ we get: $$x^{2} + y^{2} + e^{2}a^{2} = e^{2}x^{2} + a^{2}$$What happens to this equation when $x$ is replaced by $tx$ and $y$ by $ty$? In other words, how do we maintain equality of both sides when $(x, y)$ is replaced by $(tx, ty)$?

• (a) Recall that an ellipse symmetric about the origin is the set of $X$ satisfying the equation $$\|X\|^{2} + e^{2}a^{2} = e^{2}(X \cdot N)^{2} + a^{2}$$where $e$ is the eccentricity, $a = \frac{ed}{1 - e^{2}},$ and $d$ is the distance from the focus $F = eaN$ to a directrix $L,$ and $N$ is the unit normal to $L.$ Setting $X = (x, y)$ and $N = \textbf{i},$ we get $$x^{2} + y^{2} + e^{2}a^{2} = e^{2}x^{2} + a^{2}$$This means that $e^{2} = 1 + \frac{y^{2}}{x^{2} - a^{2}}.$ Rearranging terms in the above equation, we get the following:$$x^{2} + y^{2} - e^{2}x^{2} = a^{2} - e^{2}a^{2}$$Now, suppose that we transform $X$ by a scalar $t,$ in other words, $(x, y)$ becomes $(tx, ty).$ The left-hand side of the above equation then becomes $t^{2}(x^{2} + y^{2} - e^{2}x^{2}).$ But to maintain equality, we must also multiply the right-hand side by $t^{2}.$ However, solving for $e^{2},$ we see that $$\displaylines{e^{2} = \frac{t^{2}(x^{2} - a^{2} + y^{2})}{t^{2}(x^{2} - a^{2})}\\= 1 + \frac{y^{2}}{x^{2} - a^{2}}}$$Thus proving that transforming $X$ by a scalar $t$ leaves the eccentricity unchanged. This implies that scaling $X$ requires that $d$ be transformed by the same scalar.

  (b) Suppose for simplicity that the ellipses are concentric at the origin. Then, their major axes are the segments between $\pm a_{0}N$ and $\pm a_1{N}$ respectively, where $$a_{0} = \frac{ed_{0}}{1 - e^{2}}; \quad a_{1} = \frac{ed_{1}}{1 - e^{2}}$$and where $d_{0}$ and $d_{1}$ are the distances between the foci of the respective ellipses and their corresponding directrices. Since $e$ is the same value in both $a_{0}$ and $a_{1}$ we see that $$\frac{a_{0}(1 - e^{2})}{d_{0}} = \frac{a_{1}(1 - e^{2})}{d_{1}}$$Which implies that $a_{1} = \frac{a_{0}}{t}$ where $t = \frac{d_{0}}{d_{1}}.$ The standard form equation for the ellipse with major axis $\pm a_{1}N$ is $$\frac{x^{2}}{a_{1}^{2}} + \frac{y^{2}}{a_{1}^{2}(1 - e^{2})} = 1$$ But this is equivalent to $$\frac{t^{2}x^{2}}{a_{0}^{2}} + \frac{t^{2}y^{2}}{a_{0}^{2}(1 - e^{2})} = 1$$which is a scalar transform of the ellipse with major axis $\pm a_{0}N \quad \blacksquare$

  (c) The proofs of (a) and (b) are valid for conic sections with $e \neq 0.$