A problem from Spivak's Calculus:
Sketch the graph of the lemniscate $$r^2 = 2a^2\cos 2 \theta.$$
- Find an equation for its cartesian coordinates.
- Show that it is the collection of all points satisfying $d(p, (a,0))d(p, (-a,0)) = a^2$.
- Make a guess about the shape of the curves formed by the set of all $P$ satisfying $d(p, (a,0))d(p, (-a,0)) = b$, when $b<a^2$
and $b>a^2$.
For the equation in cartesian coordinates, I get $(x^2+y^2)^2 = 4a^2xy$. But when I simplify $\sqrt{(x-a)^2 + y^2}\sqrt{(x+a)^2 + y^2}=a^2$, I get $$ (x^2-2xa+a^2+y^2)(x^2+2xa+a^2+y^2)=a^4\\ \iff (x^2+y^2)^2-4x^2a^2 +2a^2(x^2+y^2)+a^4 = a^4\\ \iff (x^2+y^2)^2 = 2a^2(x^2-y^2), $$ which seems different. What am I doing wrong?
$\endgroup$2 Answers
$\begingroup$$$\cos(2\theta)=\cos^2\theta-\sin^2\theta=(x/r)^2-(y/r)^2$$
You seemed to mistake $\cos(2\theta)$ with $\sin(2\theta)$ and assume that:
$$\cos(2\theta)\not=2\cos\theta\sin\theta=2(x/r)(y/r)$$
$\endgroup$ 1 $\begingroup$I don't know where you got the equation $(x^2+y^2)^2=4a^2xy$, but that's wrong.
To find the equation in Cartesian coordinates,
$$r^2=2a^2\cos{2\theta}\\ \implies r^4=2a^2r^2\cos{2\theta}\\ \implies (x^2+y^2)^2=2a^2r^2(\cos^2{\theta}-\sin^2{\theta})=2a^2(x^2-y^2)$$
$\endgroup$
