- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
13.25. Miscellaneous review exercises on conic sections
- Use the Cartesian equation which represents all conics of eccentricity $e$ and center at the origin to prove that these conics are integral curves of the differential equation $y' = (e^{2} - 1)x/y.$ Note: Since this is a homogeneous differential equation (Section 8.25), the set of all such conics of eccentricity $e$ is invariant under a similarity transformation. (Compare with exercise 13.)
- The Cartesian equation is $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}(1 - e^{2})} = 1$$What is its derivative?
- Taking the derivative of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}(1 - e^{2})} = 1$ with respect to $x$ we get $$\frac{2x}{a^{2}} + \frac{2y(y')}{a^{2}(1 - e^{2})} = 0$$Solving for $y',$ we get $$y' = (e^{2} - 1)\frac{x}{y}$$But this implies that the set of Cartesian equations for a conic of eccentricity $e$ and center at the origin is the set of indefinite integrals of this function. $\blacksquare$