Question

Except for degenerate cases, the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$defines $a\left(n\right)\_\_\_\_\_\_\_\_\_\_$ if $B^2-4AC=0$.

Short answer

Except for degenerate cases, the equation$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$defines a parabola if$B^2-4AC=0$.

Step-by-step solution

Step 1. Given information

An equation is$A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$.

Step 2. What does an equation Ax2+Bxy+Cy2+Dx+Ey+F=0 represents?

If $B^2-4AC=0$, then either$B^2=0\text{or}4AC=0\text{,}$

Now it's not a case of degenerate axis, so $B=0$here, or

$A\ne 0\text{or}C\ne 0$

but not all three simultaneously, so the form of given equation is either

$\begin{array}{rcl}A{x}^2+Dx+Ey+F& =& 0,A\ne 0\\ C{y}^2+Dx+Ey+F& =& 0,C\ne 0\end{array}$

$\begin{array}{rcl}A\left(x^2+2\times \frac{1}{2}\times \frac{D}{A}x\right)+Ey+F& =& 0\\ A\left(x^2+2\times \frac{{\displaystyle 1}}{{\displaystyle 2}}\times \frac{{\displaystyle D}}{{\displaystyle A}}x+\frac{{\displaystyle D^2}}{{\displaystyle 4A^2}}\right)+Ey-F& =& 0\\ A{\left(x+\frac{{\displaystyle D}}{{\displaystyle 2A}}\right)}^2& =& -Ey+F+\frac{{\displaystyle D^2}}{{\displaystyle 4A^2}}\end{array}$

On dividing both sides by $A$,

$\begin{array}{rcl}\frac{A{\left(x+\frac{D}{2A}\right)}^2}{A}& =& \frac{1}{A}\left(-Ey+F+\frac{D^2}{4A^2}\right)\\ {\left(x+\frac{D}{2A}\right)}^2& =& -\frac{E}{A}\left(y-\frac{F}{E}-\frac{D^2}{4E{A}^2}\right)\end{array}$

$(x-h{)}^2=-4a(y-k)$

Similarly,

$C{y}^2+Dx+Ey+F=0C\ne 0,$denotes a parabola.

Thus except for degenerate cases, the equation $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$defines a parabola, if$B^2-4AC=0$