How do you name the curve given by the conic

Convert to the General Cartesian Form:
${\displaystyle A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0}$
Compute the determinant:
${\displaystyle \mathrm{\Delta }=B^2-4AC}$
If ${\displaystyle \mathrm{\Delta }<0}$, then it is an ellipse or a circle. If ${\displaystyle B=0\text{and}A=C}$, then it is a circle. Otherwise, it is an ellipse.
If ${\displaystyle \mathrm{\Delta }=0}$, then it is a parabola.
If ${\displaystyle \mathrm{\Delta }>0}$, them it is a hyperbola.
Given: ${\displaystyle r=\frac{4}{1+\cos \left(\theta \right)}}$
${\displaystyle r+r\cos \left(\theta \right)=4}$
Substitute ${\displaystyle r=\sqrt{x^2+y^2}\text{and}r\cos \left(\theta \right)=x}$:
${\displaystyle \sqrt{x^2+y^2}+x=4}$
${\displaystyle \sqrt{x^2+y^2}=4-x}$
${\displaystyle x^2+y^2=x^2-8x+16}$

${\displaystyle y^2+8x-16=0}$
Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, ${\displaystyle A=B=E=0,C=1,D=8,\text{and}F=-16}$
${\displaystyle \mathrm{\Delta }=0^2-4\left(0\right)\left(1\right)=0}$
It is a parabola.