Parallel and perpendicular, part 2

Calculus Level 4

{ x 2 + y 2 = 1 x 2 − 2 x y + 3 y 2 = k \begin{cases} x^2+y^2=1 \\ x^2-2xy+3y^2 =k\end{cases}

The above two curves touch each other at k = a k=a and k = b k=b where a > b a>b .

Let the area enclosed between the two curve be M M and m m for the case k = a k=a and k = b k=b respecively. Find the value of ⌊ 1000 M m ⌋ \left\lfloor{\dfrac{1000M}{m}}\right\rfloor .


The answer is 2414.

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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1 solution

Otto Bretscher
Nov 22, 2015

The symmetric matrix of the quadratic form q ( x , y ) = x 2 − 2 x y + 3 y 2 q(x,y)=x^2-2xy+3y^2 is A = [ 1 − 1 − 1 3 ] A=\begin{bmatrix}1&-1\\-1&3\end{bmatrix} and the area of the ellipse q ( x , y ) = k q(x,y)=k is k π det ⁡ A = k π 2 \frac{k\pi}{\sqrt{\det A}}=\frac{k\pi}{\sqrt{2}} . Using the values k = 2 ± 2 k=2\pm \sqrt{2} from the first part , we find M = 2 π , m = ( 2 − 2 ) π M=\sqrt{2}\pi, m=(2-\sqrt{2})\pi and M m = 1 + 2 ≈ 2.414 \frac{M}{m}=1+\sqrt{2}\approx 2.414 . The answer we seek is 2414 \boxed{2414}

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Moderator note:

Great! An understanding of quadratic forms and discriminants provide a further understanding that is useful for solving otherwise tedious problems like this.

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