What is the maximum value of $x^2+4xy-y^2$ for all $(x,y)$ satisfying $x^2+y^2 = 1$?

Question

Does the trick have something to do with the equation of a circle?

Answer 1

Hint. Let $x=\sin\theta, y=\cos\theta$ to have $x^2+4xy-y^2=\sin^2\theta+4\sin\theta\cos\theta-\cos^2\theta$. Can you use calculus to minimise this function of one variable?

Answer 2

Let $x^2+4xy-y^2=c$ and $\dfrac c1=\dfrac{x^2+4xy-y^2}{x^2+y^2}$

$\iff(c-1)(x/y)^2-4(x/y)+c+1=0$

As $x/y$ is real,the discriminant must be $\ge0$

$$4^2\ge4(c^2-1)\iff c^2\le5$$

The equality will occur if $\dfrac xy=\dfrac4{2(c-1)}$

Alternatively,

Let $x^2+4xy-y^2=u\implies x^2-y^2=u-4xy$

$1^2=(x^2-y^2)^2+(2xy)^2=(u-4xy)^2+(2xy)^2\iff20(xy)^2-8u(xy)+u^2-1=0$

The discriminant $(8u)^2-80(u^2-1)=16(5-u^2)$