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

Answer: 6

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