Why is the graph of $8x^2+8xy+2y^2+2x+y+5=0$ empty?

Question

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.

Answer 1

By the usual checks as you mention it is a pair of straight lines. But under what conditions? The conditions should be gleaned from 3D.

Conic sections arise as special cases of three dimensional second degree non-central conicoids intersected by a plane.

The parabolic cylinder ( because intersections are a pair of straight lines and discriminant $( 8^2- 4 8 2 =0 $).

In 3 dimensional situation..

$$ z= 8x^2+8xy+2y^2+2x+y+p $$

does not always cut $xy$ plane $(z=0)$ in real points.

A factoring suggested by Oscar Lanzi below for the quadratic in $(2x+y)$

$$ 2 ( 2x+y)^2 + (2x +y) +p = 0 $$

will have coincident roots when discriminant vanishes or,

$$ 1^2-4\cdot 2\cdot p =0 , i.e., when \quad p=\frac18$$

So the real pair of straight line projections in xy plane occur only when

$$p < \frac18, $$

real coincident pair of straight lines occur when the cylinder is touched by plane $z=0$ when

$$ p = \frac18$$

and none for

$$ p > \frac18 $$

For $p=5$ no real, but imaginary intersections !

EDIT1:

The equations of the pair of straight lines are accordingly give by:

$$ \dfrac{4(2 x +y) +1 }{\sqrt8}= \pm\sqrt{\frac18-p}$$

Geometrical interpretation

Reducing the pair to normal / polar angle form $( x \cos\alpha+y\sin \alpha=p),$ we have respectively for the three cases.. real pedal distances $p$ from the origin, pass through origin and have imaginary distance to the imaginary pair of straight lines.

This would be no more odd than saying the circle $ x^2+ y^2 = -16\,$ has $\,4i$ imaginary units radius.

We can reduce your question to the simplest terms:

Why is my graph of line pair

$$(x + y)^2 + 2= 0 $$ empty?

Because this pair of imaginary lines are distanced $\pm i$ from the origin,

whereas $(x + y)^2 - 2= 0 $ are a pair of real straight lines distanced at $\pm 1 $ from origin along pedal normals.

Answer 2

$$ 8 \left(x + \frac{y}{2} + \frac{1}{8} \right)^2 + \frac{39}{8} $$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 8 } & 0 & 1 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 8 & 2 \\ 8 & 4 & 1 \\ 2 & 1 & 10 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 8 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & \frac{ 39 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 0 & 1 \\ \frac{ 1 }{ 8 } & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 16 & 0 & 0 \\ 0 & \frac{ 39 }{ 4 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 8 } \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 16 & 8 & 2 \\ 8 & 4 & 1 \\ 2 & 1 & 10 \\ \end{array} \right) $$

I think the (real) solution set is empty. Back in a few minutes

Answer 3

Let's see.

$2(2x+y)^2+(2x+y)+5=0$

$2z^2+z+5=0$ where $z=2x+y$.

But the discriminant is negative -> no real $z$ -> no real $(x,y)$.

When both of your quantities are zero you guarantee two straight lines, which may be either parallel in real space, coincident or parallel in complex space (the latter meaning you get nothing in real coordinates). You can't distinguish between these possibilities with just the two discriminants alone, because they are both unchanged by changing the value of $c$ even though the graph does change with $c$. The 3×3 determinant fails to adjust with $c$ when $h^2-ab=0$ because the minor expansion in the third row gives the term $-c(h^2-ab)$. You need an additional discriminant for this case.

Render

$(\text{coefficient of }x)^2-4(\text{coefficient of }x^2)(\text{constant})$

or

$(\text{coefficient of }y)^2-4(\text{coefficient of }y^2)(\text{constant})$

which in this case will both have the same sign. If both are positive the lines are parallel in real space, if both zero they are coincident, if both negative they disappear into complex space. I leave the calculation in this case to you.

Answer 4

(as Oscar Lanzi started...)

\begin{align} 2(2x+y)^2 + (2x+y) + 5 &= 0 \\ 16(2x+y)^2 + 8(2x+y) &= -40 \\ [4(2x+y)+1]^2 &= -39 \\ \end{align}