Question

If $2x=t+\sqrt{t^2+4}$ and $3y=t-\sqrt{t^2+4}$ then the value of $y$ when $x=\frac{2}{3}$, is ____.

• A. $-2$
• B. $1$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

Let $2x=t+\sqrt{t^2+4}$ ....(1)

$3y=t-\sqrt{t^2+4}$ .....(2)

Adding eqn (1) and (2), we get

$2x+3y=2t$ .....(3)

On substracting, we get

$2x-3y=2\sqrt{t^2+4}=2\sqrt{(\frac{2x+3y}{2}{)}^2+4}$ .... From (3)

$\Rightarrow (\frac{2x-3y}{2}{)}^2=(\frac{2x+3y}{2}{)}^2+4$

$\Rightarrow -3xy=3xy+46xy=-4$

If $x=\frac{2}{3}$, then $y=\frac{-4\times 3}{2\times 6}=-1$