- 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
- Find all positive numbers $A$ and $B,$ $A\ > B,$ such that the area of the region enclosed by the ellipse $Ax^{2} + By^{2} = 3$ is equal to the area of the region enclosed by the ellipse $$ \begin{align*} \\ (A + B)x^{2} + (A - B)y^{2} = 3 \end{align*} $$
- Rewrite the equations $Ax^{2} + By^{2} = 3$ and $(A + B)x^{2} + (A - B)y^{2} = 3$ as $$\frac{x^{2}}{a_0^{2}} + \frac{y^{2}}{b_0^{2}} = 1; \quad \frac{x^{2}}{a_1^{2}} + \frac{y^{2}}{b_1^{2}} = 1$$respectively, where $$a_{0}^{2} = \frac{3}{A}, \ b_{0}^{2} = \frac{3}{B}; \quad a_{1}^{2} = \frac{3}{(A + B)}, \ b_{1}^{2} = \frac{3}{(A-B)}$$then, use the result from problem (1) to find the areas bounded by the two ellipses.
- We know from the proof of problem (1) that an ellipse with the standard form $$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$$has an area $ab$ times that of a circle with radius 1. As such, the area of the region enclosed by the ellipse $Ax^{2} + By^{2} = 3$ is $\frac{3}{\sqrt{AB}}$ and the area of the ellipse enclosed by $$(A + B)x^{2} + (A - B)y^{2} = 3$$ is $\frac{3}{\sqrt{(A+B)(A-B)}}.$ Setting these two areas equal to one-another, we see $$\frac{3}{\sqrt{AB}} = \frac{3}{\sqrt{(A+B)(A-B)}} \quad (\text{where} \ AB>0,\ A^{2} - B^{2}> 0)$$We can write this in the equivalent form $$A^{2} - AB - B^{2} = 0 \quad (\text{where} \ AB>0, \ A^{2} - B^{2}> 0)$$Solving this quadratic equation with $a = 1, b = -B, c = -B^{2}$ we find that $$\displaylines{A &= \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \\&=\frac{(1 \pm \sqrt{5})B}{2}}$$But we know that $A \ > B, \ AB \ > 0, \ A^{2} - B^{2} > 0,$ thus $$A = \frac{(1 + \sqrt{5})B}{2} \quad (\text{where}\ B > 0)\quad \blacksquare$$