The End of Faith in Quadratics

Alright, we start from the expression $S$ . $S=\frac{x^2+y^2}{ay^2+bxy+cx^2}$ Now maximum and minimum values of $S$ will be same as that of $S'=\frac{x^2+y^2}{ax^2+bxy+cy^2} \quad (Why)$

Now putting, $x=r\cos \theta , y=r \sin \theta$ we finally get, $S'= \frac{2}{(a+c)+(c-a) \cos 2\theta+b \sin 2\theta}$ Now we know that, $-\sqrt{A^2+B^2}\leq (A\sin X+B \cos X) \leq \sqrt{A^2+B^2} \space (why)$ Hence using this and further solving, we can say that the minimum value $(m)$ and maximum value $(M)$ of $S'$ will be $m=\frac{2}{(a+c)+\sqrt{(c-a)^2+b^2}}$ $M=\frac{2}{(a+c)-\sqrt{(c-a)^2+b^2}}$

Now after lots of further solving we finally get $\Rightarrow(m+M)=\frac{4(a+c)}{4ac-b^2}$

Now we know that minimum value of the quadratic expression $ax^2+bx+c$ will occur for $a>0$ and will be equal to $m_q=\frac{4ac-b^2}{4a}$

Rearranging we get $(M+m)=\frac{1}{m_q}(1+\frac{c}{a})= \frac{1}{6}(1+53)=\boxed 9$

I encourage you to add your solution for the sake of variety

Since, $S=\frac{x^2+y^2}{ay^2+bxy+cx^2}$ is a homogeneous equation in $x$ and $y$ , So, we can write , $y=mx$ [where, m is a arbitrary constant] Putting it in $S$ we get, $S=\frac{m^2+1}{am^2+bm+c}$ $aSm^2 + bSm + cS = m^2+1$ $(aS-1)m^2+bSm+cS-1=0$ Since $m$ is real, $discriminant\geq0$ $(b^2-4ac)S^2+4(c+a)S-4\geq0 ..............(i)$ As given above that, minimum value of $ax^2+bx+c$ is $6$

$i.e.$ $\frac{4ac-b^2}{4a}=6$ Here we get that $b^2-4ac=-24a$ also we are given that $\frac{c}{a}=53\space i.e. c=53a$ Putting these values in $(i)$ , we get $-24aS^2+216aS-4\geq0$ $6aS^2-54aS+1\leq0$ I used wolfram alpha to find the interval of S, Then adding up the extremities got $\frac{1}{6}(27+27)=\boxed9$