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

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Answer 1 · 2018-01-13T19:23:59

$Ax^2+Bxy+Cy^2+Dx+Ey+F = 0$

Compute the determinant:

$Delta = B^2-4AC$

If $Delta <0$ , then it is an ellipse or a circle. If $B = 0 and A=C$ , 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: $r=4/(1+cos(theta))$

$r+rcos(theta) = 4$

Substitute $r = sqrt(x^2+y^2)$ and $rcos(theta) = x$ :

$sqrt(x^2+y^2)+x = 4$

$sqrt(x^2+y^2) = 4-x$

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

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

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

$Delta = 0^2-4(0)(1) = 0$

It is a parabola.

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