Question

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

• A. 3
• B. 5
• C. -5
• D. -3

Answer 1

Answer: (A)

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