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

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Douglas K.
Jan 13, 2018

Convert to the General Cartesian Form :

#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.

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