How do you name the curve given by the conic r=4/(1+costheta)?

Jan 13, 2018

Convert to the General Cartesian Form :

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

Compute the determinant:

$\Delta = {B}^{2} - 4 A C$

If $\Delta < 0$, then it is an ellipse or a circle. If $B = 0 \mathmr{and} A = C$, then it is a circle. Otherwise, it is an ellipse.

If $\Delta = 0$, then it is a parabola.

If $\Delta > 0$, them it is a hyperbola.

Given: $r = \frac{4}{1 + \cos \left(\theta\right)}$

$r + r \cos \left(\theta\right) = 4$

Substitute $r = \sqrt{{x}^{2} + {y}^{2}}$ and $r \cos \left(\theta\right) = x$:

$\sqrt{{x}^{2} + {y}^{2}} + x = 4$

$\sqrt{{x}^{2} + {y}^{2}} = 4 - x$

${x}^{2} + {y}^{2} = {x}^{2} - 8 x + 16$

${y}^{2} + 8 x - 16 = 0$

Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, $A = B = E = 0 , C = 1 , D = 8 , \mathmr{and} F = - 16$

$\Delta = {0}^{2} - 4 \left(0\right) \left(1\right) = 0$

It is a parabola.