Question

If the equation $2x^2+4xy+7y^2-12x-2y+t=0$ where ' $t$ ' is a parameter has exactly one real solution of the form $(x,y)$. Then the $\mathrm{sum}$ of $(x+y)$ is equal to

• A. 3
• B. 5
• C. -5
• D. -3

Answer 1

Answer: (A)

$2x^2+4x(y-3)+7y^2-2y+t=0$
$D=0\text{ (for one solution) }$
$\Rightarrow 16(y-3{)}^2-8\left(7y^2-2y+t\right)=0$
$\Rightarrow 2(y-3{)}^2-\left(7y^2-2y+t\right)=0$
$\Rightarrow 2\left(y^2-6y+9\right)-\left(7y^2-2y+t\right)=0$ $\Rightarrow -5y^2-10y+18-t=0$
$\Rightarrow 5y^2+10y+t-18=0$
$\text{ again }D=0\text{ (for one solution) }$
$\Rightarrow 100-20(t-18)=0\Rightarrow 5-t+18=0$
$\therefore t=23$
$\text{ for }t=23;5y^2+10y+5=0$
$(y+1{)}^2=0\Rightarrow y=-1$
$\text{ for }y=-1;2x^2-8x+32=0$
$x^2-4x+16=0$
$x=4\Rightarrow x+y=3$