Mathematical Immaturity

13.25. Miscellaneous review exercises on conic sections

• (a) Prove that the Cartesian equation \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1 \end{align*} represents all conics symmetric about the origin with foci at $(c, 0)$ and $(-c, 0).$
  (b) Keep $c$ fixed and let $S$ denote the set of all such conics obtained as $a^2$ varies over all positive numbers $> c^2.$ Prove that every curve in $S$ satisfies the differential equation \begin{align*} xy\left(\frac{dy}{dx}\right)^2 + (x^2 - y^2 - c^2)\frac{dy}{dx} - xy = 0. \end{align*}
  (c) Prove that $S$ is self-orthogonal; that is, the set of all orthogonal trajectories of curves in $S$ is $S$ itself. [Hint: Replace $y'$ by $-1/y'$ in the differential equation in (b).]
• (a) Turn the Cartesian equation provided into a standard form equation. What is $e^2$ in this case?
  (b) The right-hand side of the derivative of the Cartesian equation in (a) is 0, this allows for multiplication of both sides by different values while maintaining equality.
  (c) Perform the subsitution suggested by the hint in the problem and then multiply both sides by $-(y')^2.$
• (a) If we set $e^2 = \frac{c^2}{a^2},$ then the Cartesian equation becomes $$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$$Simplifying terms, we get $$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$$If we denote $X = (x, y)$ and let $N = \textbf{i},$ then rearranging terms gives us $$\|X\|^2 + e^2a^2 = e^2(X \cdot N)^2 + a^2$$But if $F = c = eaN$ where $a = \frac{ed}{1 - e^2}$ with $d$ being the distance from $F$ to the directrix line $L,$ then the set of points satisfying the Cartesian equation $$\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$$is a conic section with eccentricity $e \neq 1,$ symmetric about the origin as per Theorem 13.19 $\quad \blacksquare$
  
  (b) First, we take the derivative with respect to $x$ of the Cartesian equation in (a): $$\frac{2x}{a^2} + \frac{2yy'}{a^2 - c^2} = 0$$This simplifies to $$a^2(x + yy') - c^2 = 0$$Which we can then rewrite as $$\left(\frac{a^2 - c^2}{a^2}\right)x + yy' = 0$$This equation remains unchanged when multipliying both sides by $x:$ $$\left(\frac{a^2 - c^2}{a^2}\right)x^2 + xyy' = 0$$But from the Cartesian equation we know that $\left(\frac{a^2 - c^2}{a^2}\right)x^2 = (a^2 - c^2 - y^2),$ giving us $$(a^2 - c^2 - y^2) + xyy' = 0$$Again we use the Cartesian equation to show that $a^2 = x^2 + \frac{a^2y^2}{a^2 - c^2},$ giving us $$\left(x^2 + \frac{a^2y^2}{a^2 - c^2}\right) - c^2 - y^2 + xyy' = 0$$Multiplying both sides by $y'$ and rearranging terms, we get $$xy(y')^2 +(x^2 - y^2 - c^2)y' + \frac{a^2y^2y'}{a^2 - c^2} = 0$$But we know from the derivative of the Cartesian equation that $x = \frac{-a^2yy'}{a^2-c^2},$ turning the above equation (substituting the Leibniz notation $\frac{dy}{dx}$ for $y'$) into: $$xy\left(\frac{dy}{dx}\right)^2 + (x^2 - y^2 - c^2)\frac{dy}{dx} - xy = 0 \quad \blacksquare$$ \begin{align*} \end{align*} (c) If we replace $y'$ with $-\frac{1}{y'}$ in the equation from part (b), we get $$\frac{xy}{(y')^2} - \frac{(x^2 - y^2 - c^2)}{y'} - xy = 0$$Multiplying both sides by $-(y')^2,$ we get $$xy(y')^2 + (x^2 - y^2 - c^2)y' - xy = 0$$Thus showing that any curve in $S$ is self-orthogonal as the differential equation is satisfied equally by $y'$ and $-1/y' \quad \blacksquare$