Tricky maxima minima

Let $x^{2} + y^{2} = a^{2}$ ( a varies as (x,y varies)

$\frac{x}{a} = cosz , \frac{y}{a} = sinz$

$\huge{\frac{ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}}{ \frac{x^{2}}{a^{2}} + \frac{x}{a}.\frac{y}{a} + 4\frac{y^{2}}{a^{2}}}}$

$\huge{ \frac{1}{cos^{2}z + sinzcosz + 4sin^{2}z}}$

$\huge{ = \frac{2}{1 + cos2z + sin2z + 4(1 - cos2z)}}$

$f(z) =\frac{2}{5 + sin2z + 3cos2z}$

$f(z)_{min} = \frac{2}{5 + \sqrt{10}} = \frac{2}{15}( 5 - \sqrt{10})$

$f(z)_{max} = \frac{2}{5 - \sqrt{10}} = \frac{2}{15}(5 + \sqrt{10})$

$A = \frac{2}{3} , 6A =4$

It can be done by using quadratic also.

Let the given expression be equal to t. On simplification (t - 1) x^2 + t xy + (4t - 1) = 0 has to have non-negative discriminant for real values of x and y. For -15t^2 + 20 t - 4 to be non-negative the two extreme values of t, the roots add to 20/15, i.e. 4/3, thus average 2/3 and multiplied by 6 gives 4 as answer.