Question

If $x$ and $y$ be real , and $9x^2+2xy+y^2-92x-20y+244=0$, then the value of $y$ can be

• A. $1$
• B. $2$
• C. $6$
• D. All of the above

Answer 1

The correct option is D All of the above

Given equation is $9x^2+2xy+y^2-92x-20y+244=0$

$\Rightarrow 9x^2+(2y-9)x+y^2-20y+244=0$

This is quadratic in $x$.

If $x$ and $y$ are real, then $D\ge 0$

Thus ${(2y-92)}^2-4\times 9(y^2-20y+244)\ge 0$

$\Rightarrow 4y^2+8464-368y-36y^2+720y-8784\ge 0\Rightarrow -32y^2+352y-320\ge 0\Rightarrow y^2-11y+10\le 0\Rightarrow y^2-10y-y+10\le 0\Rightarrow (y-10)(y-1)\le 0\Rightarrow y\in [1,10]$

All the options are in the desired range.

So, option E is correct