$ax^2 + 2hxy + by^2 +2gx + 2fy+c = 0$ represents a point when $\Delta = 0$ and $h^2 - ab < 0$. How can I show that $ax^2  + 2hxy + by^2 +2gx + 2fy+c = 0$  represents a point when  $\Delta = 0$  and $h^2 - ab < 0$.
I know I can do it by completing the square. But it turns out to be too tedious. Can anyone help me ?
 A: From the Wikipedia page on Degenerate conic - Discriminant, we have that:
The conic $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a point (a degenerate case of an ellipse) when the matrix
$$M_1 = \begin{bmatrix}a & h & g\\h & b & f \\ g & f & c\end{bmatrix}$$
has determinant $\det(M_1)=0$, and when the matrix
$$M_2 = \begin{bmatrix}a & h \\h & b \end{bmatrix}$$
has determinant $\det(M_2)>0$.
You can double-check and verify that these determinantal constraints give your required conditions.
A: We can write $$y=\frac{hx+f\pm \sqrt{(h^2-ab)x^2+2(hf-bg)x+(f^2-ac)}}{b}~~~~~(1)$$
For a pair of lines, $y$ needs to be real for all real values of $x$. inside square root we should have a perfect square of like $(px+q)^2>0$. For  a point it should be like $-(rx+s)^2$ ($y$ non real excepting one value of $x=-s/r$. For this we need to have  $A<0$ and $B^2-4AC=0$ for the quadratic in the radical. The two conditions give  $h^2-ab<0, (hf-bg)^2-(h^2-ab)(f^2-ac)=0=\Delta.~~~~(2)$
The examples are: $x^2+y^2=0, 3(x-1)^2+(y-x+2)^2=0, 3(x-1)^2+2(y-2)^2=0.$
Note that if $x,y$ are real then sum of perfect squares cannot be zero. The only possibility is that both squares are simultaneously zero. the points are $(0,0),(1,-1),(1,2)$ are the points represented by the three examples, respectively. Upon expansion, for them we get $a,b,c,f,g,h$ satisfying both parts of (2).
