Question

If $x,y\in R$ and $2x^2+6xy+5y^2=1,$ then

• A. $|x|\le \sqrt{5}$
• B. $|x|\ge \sqrt{5}$
• C. $y^2\le 2$
• D. $y^2\le 4$

Answer 1

Answer: (A), (C)

The correct options are
A $|x|\le \sqrt{5}$
C $y^2\le 2$

Given, $2x^2+6xy+5y^2=1$ ......... (1)

It can be rewritten as

$2x^2+\left(6y\right)x+5y^2-1=0$

since $x$ is real

$D\ge 0$

$36y^2-8(5y^2-1)\ge 0$

$\Rightarrow y^2\le 2$

$\Rightarrow -\sqrt{2}\le y\le \sqrt{2}$

Equation (1) can also be rewritten as

$5y^2+\left(6x\right)y+2x^2-1=0$

since $y$ is real

$D\ge 0$

$36x^2-20(2x^2-1)\ge 0$

$\Rightarrow 36x^2-40x^2+20\ge 0$

$\Rightarrow -4x^2\ge -20$

$\Rightarrow x^2\le 5$

$\Rightarrow -\sqrt{5}\le x\le \sqrt{5}$

$|x|\le \sqrt{5}$

Option A and C are correct.