# How do you name the curve given by the conic

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

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