Question

If $M$ and $m$ are the maximum and minimum values of $\frac{y}{x}$ for pair of real numbers $\left(x , y\right)$ which satisfy the equation $\left(x - 3\right)^{2}+\left(y - 3\right)^{2}=6$ , then the value of $\frac{1}{M}+\frac{1}{m}$ is

Answer 1

Let $\frac{y}{x}=m^{'}$ . So, on putting $y=m^{'}x$ in the given circle,
We get $\left(x - 3\right)^{2}+\left(m^{'} x - 3\right)^{2}=6$
$\Rightarrow x^{2}\left(1 + m^{'}\right)-6x\left(1 + m^{'}\right)+12=0$
Putting discriminant $=0$ , we get,
$m^{\prime}=3 \pm 2 \sqrt{2}$
So, $m=3-2\sqrt{2}$ and $M=3+2\sqrt{2 }\Rightarrow mM=1$
$\Rightarrow \frac{1}{M}+\frac{1}{m}=m+M$
$=3-2\sqrt{2}+3+2\sqrt{2}$
$=6$