Question: An elipse with equation $$ {x^2\over a^2} + {y^2\over b^2} = 1 $$ is enclosed by the hyperbolas given by $xy=1$ and $xy=-1$. , Determine the largest area of an ellipse enclosed by the hyperbolas. Note that the area of the ellipse is $$ A = πab$$
I know that the answer is $A=2\pi$ by solving it this way:
Considering the hyperbola $y=\frac{1}{x} \Leftrightarrow y^2 = \frac{1}{x^2} $
Now
Ellipse is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2}= 1 $$
$$ x^2b^2 + y^2a^2-a^2b^2=0 $$
$$ x^2b^2 + \frac{a^2}{x^2}-a^2b^2=0$$
$$x^4b^2 - a^2b^2x^2+a^2=0$$
As this is a biquadratic and with the ellipse it appears to be a 4 point symmetric (i.e. tangent) hence discriminant must be equal to zero.
so
$$ b^2-4ac=0$$
$$ a^4b^4-4b^2a^2=0$$
$$ a^2b^2(a^2b^2-4)=0$$
$$ ab = 2 $$
Hence max area is $A=2\pi$
But I was wondering if there is an alternative way of finding out this area through integration or parametric equations, surely there is more than one way?
$u = \frac xa\\ v = \frac yb\\ \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \to u^2 + v^2 = 1\\ xy = 1 \to uv = \frac {1}{ab}$
$uv = k$ touches $u^2 + v^2 = 1$ when $k =\frac 12$
For any ellipse that touches the hyperbola, $ab = 2,$ and its area $= 2 \pi$