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 sum of (x+y) =?

Hello student,
Please find the answer to your question below
Given2x^2 +4xy + 7y^2-12x-2y+t=0
2x²+4x(y-3)+7y²-2y+t=0
discriminant should be=0 for one solution
so 16(y-3)²-8(7y²-2y+t)=0
on simplifying we get
5y²+10y+t-18=0
Againdiscriminant=0 for 1 solution
100-20(t-18)=0
we get t=23
For t=23 5y²+10y+5=0 which gives 5(y+1)²=0
we get y=-1
for y=-1 2x²-16x+32=0
we get x=4
so x+y=3