<s>[INST] Why is the graph of $8x^2+8xy+2y^2+2x+y+5=0$ empty? For the equation
$$8x^2+8xy+2y^2+2x+y+5=0$$
Comparing it with standard equation of conic section we get that
$h^2-ab=0$ and
$$\Delta= \left|\begin{array}{ccc} 
a & h & g \\
h & b & f \\ 
g & f & c \end{array}\right| = 0$$
So as per I know it should represent a pair of straight line which are real and coincident but after plotting it on desmos( an app to plot graphs) it showed nothing. Please help me.
 [/INST] $$  8 \left(x + \frac{y}{2} + \frac{1}{8} \right)^2  + \frac{39}{8}  $$
$$ P^T H P = D  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 -  \frac{ 1 }{ 8 }  & 0 & 1 \\ 
 -  \frac{ 1 }{ 2 }  & 1 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
16 & 8 & 2 \\ 
8 & 4 & 1 \\ 
2 & 1 & 10 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  -  \frac{ 1 }{ 8 }  &  -  \frac{ 1 }{ 2 }  \\ 
0 & 0 & 1 \\ 
0 & 1 & 0 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
16 & 0 & 0 \\ 
0 &  \frac{ 39 }{ 4 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
$$
$$ Q^T D Q = H  $$
$$\left( 
\begin{array}{rrr} 
1 & 0 & 0 \\ 
 \frac{ 1 }{ 2 }  & 0 & 1 \\ 
 \frac{ 1 }{ 8 }  & 1 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
16 & 0 & 0 \\ 
0 &  \frac{ 39 }{ 4 }  & 0 \\ 
0 & 0 & 0 \\ 
\end{array}
\right) 
\left( 
\begin{array}{rrr} 
1 &  \frac{ 1 }{ 2 }  &  \frac{ 1 }{ 8 }  \\ 
0 & 0 & 1 \\ 
0 & 1 & 0 \\ 
\end{array}
\right) 
 = \left( 
\begin{array}{rrr} 
16 & 8 & 2 \\ 
8 & 4 & 1 \\ 
2 & 1 & 10 \\ 
\end{array}
\right) 
$$
I think the (real) solution set is empty. Back in a few minutes
</s>